File pcm book.pdf - Pearland Independent School District

Pre-College Math
First Edition
C. C. Cogswell
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© Corey Cogswell, 2010
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Table of Contents
Unit I – Elementary Algebra 7
‣ Chapter I.A – Number Manipulation 9
I.A.1 – Fraction Arithmetic 10
I.A.2 – Numerical Conversions 16
I.A.3 – Exponents & Radicals 23
I.A.4 – Order of Operations (GEMA) 30
I.A.5 – Solving Multi-Step Equations & Inequalities 33
‣ Chapter I.B – Functions and Linear Equations 39
I.B.1 – Function Overview 41
I.B.2 – Graphing a Linear Function or Inequality 46
I.B.3 – x-intercepts & y-intercepts 51
I.B.4 – Linear Regression 53
I.B.5 – Parallel & Perpendicular Lines 56
‣ Chapter I.C – Quadratic Equations 59
I.C.1 – Polynomial Multiplication 61
I.C.2 – Factoring 64
I.C.3 – Solving a Quadratic Equation 67
I.C.4 – x-intercepts & y-intercepts 70
I.C.5 – Vertex Form 72
Unit II – Advanced Algebra 75
‣ Chapter II.A – More Functions 77
II.A.1 – Exponential Functions 79
II.A.2 – Logarithmic Functions 82
II.A.3 – Function Division 87
II.A.4 – Polynomial Functions 90
II.A.5 – Rational Functions 94
‣ Chapter II.B – Function Manipulation 97
II.B.1 – Composite Functions 99
II.B.2 – Inverse Functions 102
II.B.3 – External Transformations 105
II.B.4 – Internal Transformations 109
II.B.5 – Transformations Involving Absolute Value 113
‣ Chapter II.C – Systems of Equations 116
II.C.1 – Linear Systems Graphically 118
II.C.2 – Elimination Method 120
II.C.3 – Substitution Method 122
II.C.4 – Non-Linear Systems 124
II.C.5 – System Applications 126
‣ Chapter II.D – Matrices 127
II.D.1 – Matrix Addition & Scalar Multiplication 129
II.D.2 – Matrix Multiplication 131
II.D.3 – Determinants & Inverse Matrices 133
II.D.4 – Systems of Equations 137
II.D.5 – Matrix Transformation 140
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‣ Chapter II.E – Sequences and Series 143
II.E.1 – Recursive & Explicit Formulas 145
II.E.2 – Arithmetic & Geometric Sequences 148
II.E.3 – Arithmetic Series 151
II.E.4 – Geometric Series 154
II.E.5 – Applications of Sequences & Series 157
Unit III – Triangular Geometry 158
‣ Chapter III.A – Triangles 160
III.A.1 – Pythagorean Theorem, Distance, & the Circle Equation 162
III.A.2 – Sine, Cosine, & Tangent 165
III.A.3 – Special Right Triangles 169
III.A.4 – Triangle Area Formulas 172
III.A.5 – Triangle Applications 176
‣ Chapter III.B – Two-Dimensional Vectors 177
III.B.1 – Component Form, Magnitude, & Direction 179
III.B.2 – Vector Addition 182
III.B.3 – Dot Product 184
III.B.4 – Vector Projection 186
III.B.5 – Physics Applications 188
Unit IV – Trigonometry 189
‣ Chapter IV.A – The Unit Circle 191
IV.A.1 – Special Trig Ratios & the Hand Trick 193
IV.A.2 – Special Trig Ratios of Co-terminal Angles 197
IV.A.3 – Radian Conversions 199
IV.A.4 – Special Trig Ratios of Radian Measurements 201
IV.A.5 – Arc Length, Sector Area, & Rotational Motion 203
‣ Chapter IV.B – Trigonometric Functions 206
IV.B.1 – Graphing Sine & Cosine 208
IV.B.2 – Graphing Secant, Cosecant, Tangent, & Cotangent 212
IV.B.3 – Inverse Trig Functions & Initial Values 215
IV.B.4 – Arcsine, Arccosine, & Arctangent 218
IV.B.5 – Solving Simple Trigonometric Functions 221
‣ Chapter IV.C – Oblique Triangles and 3D Vectors 223
IV.C.1 – Law of Sines 225
IV.C.2 – Law of Cosines 227
IV.C.3 – Three Dimensional Vectors 229
IV.C.4 – Cross Product 231
IV.C.5 – Cross Product Applications 233
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Unit V – Combinatorics 235
‣ Chapter V.A – Counting Tools 237
V.A.1 – Factorial Arithmetic 239
V.A.2 – Counting Choices 241
V.A.3 – Rearrangements 243
V.A.4 – Permutations 245
V.A.5 – Combinations 247
‣ Chapter V.B – Probability 250
V.B.1 – Probability Conceptually 252
V.B.2 – Probability Graphically 256
V.B.3 – Probability using Permutations & Combinations 259
V.B.4 – Binomial Probability 261
V.B.5 – Mathematical Expectation 264
Unit VI – Logic
‣ Chapter VI.A – Truth Tables
VI.A.1 – Conjunction, Negation, Disjunction
VI.A.2 – Conditional Statements & Material Equivalence
VI.A.3 – Intermediate Truth Tables
VI.A.4 – Disjunctive Syllogism, Modus Ponens, Hypothetical Syllogism, Modus Tollens, &
Constructive Dilemma
VI.A.5 – DeMorgan’s Theorems, Material Equivalence, & Exportation
‣ Chapter VI.B – Symbolic Proofs
VI.B.1 – Rules of Inference
VI.B.2 – Rules of Replacement
VI.B.3 – Elementary Proofs
VI.B.4 – Intermediate Proofs
VI.B.5 – Word Problems
‣ Chapter VI.C – Trig Identities
VI.C.1 – Pythagorean Identities
VI.C.2 – Composite Argument Identities
VI.C.3 – Double Angle Identities
VI.C.4 – Half Angle Identities
VI.C.5 – Sum Product Identities
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Unit
I
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Unit
I Unit Overview
The Purpose of the Unit
Studies have shown that the human brain develops at different rates for different people. Most
studies suggest that the majority of students’ brains are not developed enough to understand the true
intricacies of Algebra until they reach the age of 17. However, the American public school system
has deemed it necessary to require Algebra 1 be taken by ninth grade when students are 14 or 15, an
age when they cannot truly understand it. Many students never get a strong foundation in Algebra,
and they feel perpetually weak in mathematics from then on.
The purpose of this unit is to reinforce that Algebraic foundation by emphasizing those skills that
actually relate to future topics, thereby making your mathematical success in this course and future
courses more attainable.
A Brief History of Algebra
Certain forms and concepts that are now considered as Algebra have been around for millennia.
Ancient Babylonian tablets dated around 1800 BCE demonstrate knowledge of quadratic equations.
However, Hellenistic mathematician Diophantus lived is often considered to be the "father of algebra".
He wrote Arithmetica around A.D. 200, a work featuring solutions to algebraic equations and general
number theory. For years this title has been contested by supporters of Persian mathematician
Muḥammad ibn Mūsā al-Ḵwārizmī, who wrote Al-Kitab al-Jabr wa-l-Muqabala (meaning "The
Compendious Book on Calculation by Completion and Balancing") in A.D. 820 on the systematic
solution of linear and quadratic equations. He is best known for founding algebra as an independent
discipline and for introducing the methods of "reduction" and "balancing" (the transposition of
subtracted terms to the other side of an equation, that is, the cancellation of like terms on opposite
sides of the equation) which was what he originally used the term “al-jabr” to refer to. The word “aljabr”
has evolved over time into its current form, “Algebra”.
Algebra was first introduced to Europe by Italian mathematician Leonardo Fibonacci in 1202, a
mathematician well-known for his Fibonacci sequence. He is also the first to bring Arabic numerals to
western society, who had been using Roman numerals for more than a thousand years. Between
1212 and 1282 Arab mathematician Abū al-Hasan ibn Alī al-Qalasādī took "the first steps toward the
introduction of algebraic symbolism." He used "short Arabic words, or just their initial letters, as
mathematical symbols."
Many mathematicians from many countries for many years have had their hand in developing what
we now refer to as Algebra, a study of mathematics so extensive, two courses are needed to
understand even its most basic of concepts.
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Chapter
I.A
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Chapter
I.A Chapter
Overview
‣ I.A.1 – Fraction Arithmetic
This section focuses on arithmetical operations, including addition, subtraction, multiplication, and
division, with fractional or “rational” numbers.
‣ I.A.2 – Fraction / Decimal / Percent Conversions
This section allows for conversions between fractions, decimals, percents, and scientific notation
without the need of a calculator.
‣ I.A.3 – Exponents and Radicals
This section explores the relationships between powers and roots, and the subtleties of negative
and fractional exponents.
‣ I.A.4 – Order of Operations (GEMA)
This section provides an alternative and more effective order of operations to PEMDAS.
‣ I.A.5 – Solving Multi-Step Equations and Inequalities
This section covers the solving of simple multi-step equations for an unknown variable by
elementary algebra.
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Section
I.A.1 Fraction
Arithmetic
The term fraction is one of the mathematical f-words (along with function and factoring) that makes
most people cringe. For ages, the way fractions are taught in elementary school has been
synonymous with pizza slices, but unfortunately the true meaning of and ability to calculate with
fractions has diminished considerably over the last few decades.
So what exactly is a fraction? Many people would say that a fraction is a part of a whole. If we cut a
pizza into 5 slices and then eat 2 of the slices, we have eaten of the pizza and there is of the pizza
left over. This is all well and good as long as the fractions are proper, but what if they’re improper?
What does mean? If we cut a pizza into 2 slices, there’s no way we can eat 5 of those slices. It’s at
this point where most peoples’ brains explode into a pizza-induced aneurism.
Some people may say that a fraction is simply a ratio comparing two numbers, like 3 apples to 5
oranges or vice versa, but this is not much help when calculating something involving arithmetic of
fractions.
So what exactly is a fraction? A fraction is simply an imperfect quotient. In other words, it is a
division problem that does not yield an integer answer. If we divide 6 by 2 we get = 3, which is an
integer. If we divide 6 by 5, we get , which cannot reduce to an integer, but who cares? We just
saved ourselves a step!
Let’s apply this new knowledge to our pizza dilemma. If we take 2 pizzas and divide evenly between
5 people, each person gets of a pizza or 2 out of 5 slices, the same answer we found before. No
big deal, right? If we take 5 pizzas and divide them evenly between 2 people, each person gets of
the pizzas, which technically is the same thing as 2.5 pizzas or assuming we cut each pizza into 2
equal pieces, each person gets 5 pieces. Regardless, the answer is no matter how we
conceptualize it.
Equivalent Fractions and Simplification
The fraction
is equivalent to because both 35 and 10 divide evenly by 5. Numerically speaking,
two fractions are equivalent if we can multiply the same number by both the numerator and
denominator of one fraction and get the other.
Mathese: = if and only if an = c and bn = d for some real number n.
Myth: It is always necessary to simplify a fraction to the point that the numerator and denominator
have no more common factors.
Fact: Simplifying a fraction is only useful when the fraction is to be used for further calculation. A final
answer need not be reduced. The understanding of equivalent fractions is most useful on multiple
choice tests, when answers may or may not be simplified.
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Fraction Addition
A long time ago in an elementary school far far away, we learned how to add fractions by finding the
least common denominator and “converting” the fractions to equivalencies with the new denominator.
This method required multiple steps, and typically most students do not respond well to complicated
processes. Therefore, many students to this day don’t know how to add fractions without a calculator.
Additionally, this method is really only useful if you must have a fully reduced fraction as your answer.
Example 1a (The old method)
3
5 + 5 20 = 3 ∙ 4
5 ∙ 4 + 5 20 = 12
20 + 5 20 = 12 + 5 = 17
20 20
Conceptually, if we take 12 pizzas and divide them equally between 20 people, and then we divide 5
more pizzas between the same 20 people, we will get the same answer if we start with 17 pizzas total
and divide them evenly between the 20 people.
Now that we know that simplification is not necessary and that all answers equivalent to the right
answer are right as well, we do not have to worry about finding a least common denominator; any
common denominator will do. To add two fractions, we can use the following formula.
a
b + c ad + bc
=
d bd
In other words, cross multiply to get the numerator, and multiply straight across to get the “common
denominator”.
Example 1b (The new method)
3
5 + 5 3(20) + 5(5)
= = 85
20 (5)(20) 100
Note:
=
, so we get the same answer either way.
Example 2 (Fraction Subtraction)
3
5 − 5 3(20) − 5(5)
= = 35
20 (5)(20) 100 = 7 20
As you can see, fraction subtraction works in exactly the same way.
Important note: The “–“ symbol in front of a fraction can be applied to either the numerator or
denominator but never both. The majority of cases require it to be in the numerator.
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Example 3 (Fraction-Integer Addition)
Note: Any integer can be converted into a fraction by writing the integer over 1.
5 + 8 3 = 5 1 + 8 3
=
5(3) + 8(1)
(1)(3)
= 23
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Important note: Mixed numbers are the bane of higher level mathematics. 5 should never be written
once you learn algebra. Technically, 5 = 5 + = by the method seen above. Fractions like
might be called improper, but mixed numbers are truly terrible.
In the case of adding three fractions, a similar method can be performed as with two fractions. The
major difference is that we cannot use the term “cross multiply”. The expanded formula is as follows:
a
b + c d + e adf + bcf + bde
=
f bdf
Here we can see that we multiply each numerator by the other two denominators and then multiply all
the denominators together to get a “common denominator”.
Example 3 (Fraction Addition of Three Fractions)
1
2 + 7 3 + 4 1(3)(9) + 7(2)(9) + 4(2)(3) 27 + 126 + 24
= = = 177
9 (2)(3)(9)
54 54
Fraction Multiplication
Multiplying fractions is extremely easy. We simply use the formula:
a b c d = ac
bd
There are times when reducing fractions makes calculating easier, but this method works regardless
of how many fractions we are multiplying together.
Dividing fractions is almost as easy. We have to utilize one extra step. We have to multiply by the
reciprocal of what we are dividing by as seen in the formula:
a
b ÷ c d = a b d c = ad
bc
Note: A reciprocal of a number n can be written as .
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Example 4 (Fraction Multiplication)
3 8 4 7 = (3)(4)
(8)(7) = 12
56 = 3 14
Or
= = if you cancel out 4 from the numerator and denominator.
Example 5 (Fraction Division)
3
8 ÷ 4 7 = 3 8 7 4 = 21
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Application
Whenever we use the word ‘of’ when dealing with fractions, we imply multiplication. So taking of
another amount means we multiply by that amount. Additionally, when answering a word problem,
leaving the answer improper is mathematically acceptable, but not always useful. For such a case, it
is best to convert the fraction to an approximate decimal equivalent, which we will get to in the next
section. For now improper fractions suffice. Never use mixed numbers.
Example 6 (Application)
Joan called in sick for work because she ate of the store’s supply of jelly beans in one night. If the
store had 189 jelly beans, how many jelly beans did Joan eat?
The ‘of’ indicates we need to multiply. (189) =
= = 108 jelly beans.
Notice: Since 189 is divisible by 7, we may prefer to divide by 7 first and then multiply by 4. We’ll
discuss this more in section I.A.4.
Example 7 (The problem with English)
Sam has baked 4 pies. His baking instructor asks Sam to divide his pies by half, so that he can take
half of all the pies home. How many pies does the instructor take home?
Most students would read this problem and, without much thought, determine that the instructor will
get two of the pies as two is one half of 4. However, careful reading of the problem reveals that Sam
was instructed to divide ‘by’ , which means we multiply by the reciprocal, 2. Therefore Sam was
actually instructed to bake an additional 4 pies, and the instructor took half of the 8 pies home, which
means the instructor took 4 pies home. Had the sentence read divide ‘in’ half, then the answer would
have been 2. When English and math are combined, things can get really tricky.
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I.A.1: Independent Practice
Add the following.
4. 5 + =
1. + =
3. + =
6. + =
2. + =
5. + =
Subtract the following.
7. − =
8. − =
9.
10. − =
11. 7 − =
− = 12. − 3 =
Multiply the following.
16. =
13. =
15. =
18. (14) =
14. =
17. (5) =
Divide the following.
22. ÷ =
19. ÷ =
21. ÷ =
24. 8 ÷ =
20. ÷ =
23. ÷ 6 =
Answer the following word problems.
25. What is of 10?
26. What is of 27?
27. What is of of 28?
28. Forty-year-old Carl lives at home with his mommy and daddy. He is required to give his
parents of his monthly income as rent. If Carl makes $1275 each month. How much does he
have left to spend after he pays rent?
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Section
I.A.2 Numerical
Conversions
A fraction is referred to as a rational number because it can be written as a ratio. Equivalently, any
number that can be written as a fraction is considered rational as well. This means all integers are
rational because we can place the integer over 1 to create a fraction equivalent. Additionally, all
terminating and repeating decimals are rational because we can rewrite them as fractions, but many
students do not know how.
Fraction Decimal
Remember in the previous section we described a fraction as a division problem. We can convert
any fraction to its decimal equivalent just by completing the division process.
Example 1
Convert to a decimal.
When dividing by a single digit, this process is fairly simple if you remember the concept of
remainders. When you divide 1 by 8, the answer is 0 with a remainder of 1. We then tack on a 0 to
the 1 and get 10. When we divide 10 by 8, we get 1 with a remainder of 2. We then tack on a 0 to
the 2 and get 20. When we divide 20 by 8 we get 2 with a remainder of 4. We then tack on a 0 to the
4 and get 40. When we divide 40 by 8 we get 5 with a remainder of 0. When we have a remainder of
0, we are done. All we have to do now is write the answers out to each division problem. Our
answers were 0, 1, 2, 5, which means = 0.125.
Example 2
Convert to a decimal.
Again we start off in much the same way. When we divide 7 by 3 we get 2 with a remainder of 1.
When we divide 10 by 3 we get 3 with a remainder of 1. When we divide 10 by 3 we get 3 with a
remainder of 1. At this point you should realize that logically, we would continue like this forever, and
there’s no chance of ever getting 0 as a remainder. We know our answer will turn out to be 2.333…
where the 3’s repeat forever. This is called a repeating decimal and will either end with a “…” to
indicate the pattern continues or it will be written as 2. 3 where the bar above the three indicates only
the 3 will repeat.
Decimal Fraction
This process is not as obvious as the previous one. In order to turn a decimal into a fraction we need
to remember the rule of 9’s and 0’s. 9 indicates repeating, and 0 indicates non-repeating. For every
number behind the decimal that repeats there will be a 9 in the denominator. For every number that
does not repeat behind the decimal there will be a 0 in the denominator.
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Example 3 (Terminating Decimal less than 1)
Convert 0.125 to a fraction.
There are three digits behind the decimal which means we need three 0’s in the denominator. This
converts to
, which reduces to because both 125 and 1000 can divide by 125.
Example 4 (Terminating Decimal greater than 1)
Convert 3.425 to a fraction.
Again there are three digits behind the decimal that do not repeat, so we need three 0’s in the
denominator. Therefore we get
=
.
Example 5 (Repeating Decimal)
Convert . 54 to a fraction.
There are two digits behind the decimal that is repeating, so we need two 9’s in the denominator.
Therefore we get
.
Curious Fact: . 9 = = 1.
This is all well in good when we get a terminating decimal and a decimal with only repeating terms. It
becomes tricky when we mix the two. The rule of 9’s and 0’s still stands. The numerator is
determined by subtracting the non-repeating part from the whole number (using only one repetition of
the repeating terms.
Example 6 (Combination Decimal)
Convert 3.45 to a fraction.
We have one non-repeating term behind the decimal (one 0) and one repeating term behind the
decimal (one 9). Our answer is determined as
=
.
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Percents
The word “percent” is made up of two parts. “Per” means to divide, and “cent” means 100. Literally,
a percent means to divide by 100.
Percents Decimals
When we are given a numerical percent, we simply need to divide that number by 100. When
working with a decimal, dividing by 100 means we must move the decimal point two places to the left.
The decimal equivalent is the true mathematical meaning of a percent, and is the value we use for
most calculations.
Example 7
Convert 34.56% to a decimal.
Moving the decimal point two places to the left, we get a decimal equivalent of .3456. Which can be
written as
=
.
To convert a percent to a fraction, we need to first convert to a decimal, otherwise we end up with a
fraction like .
, which is mathematically atrocious. Never mix decimals and fractions.
Decimals Percents
Since we divided by 100 to go from percents to decimals, it’s logical that we do the opposite to
convert back. Therefore to convert from decimals to percents , we must multiply by 100 or move the
decimal two spaces to the left. It is acceptable in this case to round the percentage to two decimal
places if necessary.
Example 8
Convert . 6 to a percentage.
Remember that . 6 is an infinite number of 6’s. So if we move the decimal over two places to the right
we end up with 66. 6%, which rounds to 66.67%. Sometimes you will see percentages and fractions
mixed to be more precise like 66 %. Because this is a mixed number, we will assume that this is
terrible.
Again, to convert from a fraction to a percent, we will want to find the decimal equivalent first. There
is one school of thought that creates a proportion like =
, but since this method does not
eliminate the need to divide a fraction, it’s unnecessary when you can divide the fraction as is, and
move the decimal point two places.
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Scientific Notation
Scientific notation is a method for rewriting very large of very small numbers as exponential numbers
with a base of ten. For instance 640,000 can be rewritten as 6.4 × 10 and 0.00054 can be rewritten
as 5.4 × 10 . How do we do this? If the number is larger than 1, the exponent is positive, whereas if
the number is smaller than 1, the exponent is negative. The coefficient has one digit to the left of the
decimal, and the numerical value of the exponent is how many places the decimal point moved.
Example 9
Convert 75,000 to scientific notation.
We move the decimal point 4 units to the left to get 7.5. Therefore the scientific notation equivalent is
7.5 × 10 .
Example 10
Convert 8.32 × 10 to its fraction equivalent.
The negative exponent indicates the number is less than one and the decimal point has moved two
places. The decimal equivalent would be 0.0832. The fraction equivalent would then result from
=
.
Note: Scientific notation we write as 3.5 × 10 will be written in a calculator as 3.5 ε 6. Although these
are equivalent, it is bad form to actually write the calculator version as our answer.
Applications of Percent
There are three main rules for working with percents.
1. Always calculate using the decimal equivalent.
2. Percent problem are like fraction problems in that the word “of” indicates multiplication.
3. A number always starts out as 100% of itself (Reminder 100% means 1.00 and “of” means
times. So this last rule translates to a number is 1 times itself)
Example 11
Rhonda’s sales were 45% of Barbara’s sales of $1256 last Tuesday. How much did Rhonda sell?
This translates to R = 45%(1256) = (. 45)(1256) = $565.20.
Note: Money, like percent, is often rounded to two decimal places.
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Percent Increase
To perform a percent increase of a number we need to add the percent to 100% before multiplying.
According to Rule 3, if a number is already at 100%, we then need to increase the percentage above
100%. Many students are confused by this concept, as they have been told 100% means everything.
However a percentage above 100% is simply an improper fraction. No need to worry.
Example 12
Rhonda was told she had to increase her sales of $565.20 by at least 20% by next week in order to
keep her job. What is the minimum she would need to sell?
As this is a percent increase, we know that next week’s total needs to be 120% of this week’s. We
get 120%(565.20) = 1.2(565.20) = $678.24.
Example 13
By what percent increase would Rhonda need to improve her sales from $565.20 to Barbara’s total of
$1256?
We just need to go backwards. We divide
. = 2. 2 = 222.22%. However 100% is included in her
original sales, so she would need a 122.22% increase in her sales.
Percent Decrease
Percent decrease works in much the same way, except that you subtract the percent from the starting
100% before multiplying.
Example 14
Rhonda is able to get her clothes at a 20% discount. If she purchases a shirt originally priced at $35,
what does she have to pay?
A discount is a decrease. So if we take 20% from 100%, she has to pay 80% of the original cost.
80%(35) = .8(35) = $28.
Example 15
Unfortunately Rhonda forgot that she has to pay a 5.6% sales tax as well. How much will she end up
having to pay?
A sales tax is a percent increase. We get . 8(1.056)(35) = $29.57.
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I.A.2 – Independent Practice
Complete the Table.
Fractions Decimals Percents Sci. Notation
1. 3
2.
4
2
3.
3
4
4.
7
29
5.
6
32
5
6. 3.45
7. 0.272727…
8. 0.45
9. 3.213
10. 1.312
11. 3.56%
12. 45.2%
13. 0.56%
14. 324%
15. 333.33%
16. 5.3245 × 10
17. 3.445 × 10
18. 4.21 × 10
19. 5.2137 × 10
20. 3. 2 × 10
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Answer the following.
21. The population is increasing in Vermont by 2.6% every year. If the population was 2,000,000
last year, what is the population this year? What will the population be next year?
22. After applying a vaccine to a virus, studies have shown a loss of 32% of the virus every two
hours. If John began with 45,000,000 virus cells in his body, how many will be left two hours
after the vaccine has been administered? How many will be left after ten hours?
23. Cindy desperately wants to buy a blouse on tax-free weekend. The original price of the blouse
is $42. It’s on sale for 25% off, and she has a coupon for 15% off the new sale price. How
much will she have to spend on her blouse?
24. Wallace would like to purchase a new game system at $299.99. He received $350 for his
birthday. With a 6.5% sales tax, can he afford his new game system? If so, what is the most
expensive game he could purchase considering sales tax?
25. Electronics Depot is selling televisions normally priced at $2499.99 for 1799.99. What percent
discount are they offering?
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Section
I.A.3 Exponents &
Radicals
It is general knowledge that many mathematicians have no life. What time they do not spend
investigating mathematics and finding more and more complex (and non-applicable) branches of
mathematics to justify their sad existence, they spend coming up with shortcuts to make their life
easier. Of course this directly contradicts the fact that they have no life, but every once in a while
mathematicians need some positive reinforcement too. The usual problem with these simplified
shortcuts is that they are usually disguised with unusual symbols that make non-mathematicians cry
themselves to sleep at night. Here are few of those symbols for those who need a good cry:
∞, ∀, ∆, ∃, ∈, √ , ∫, ∑, ∏, etc… One actually useful shortcut was multiplication.
What is multiplication? Multiplication is simply shortcut for the addition of identical numbers.
2 × 3 really means two 3’s added together or three 2’s added together. In other words
2 × 3 = 3 + 3 = 2 + 2 + 2 = 6. Additionally, we say multiplication is commutative because 2 × 3 =
3 × 2.
So how do we shortcut multiplication? What happens if we have 3 × 3 × 3 × 3 × 3? For this type of
problem we can use exponentiation. We can rewrite 3 × 3 × 3 × 3 × 3 as 3 , where the superscript
5 indicates we are multiplying five 3’s together. Exponentiation is not commutative: 3 ≠ 5 .
Reminder: In multiplication two negative symbols cancel each other out. Therefore, any even number
of negative symbols will cancel each other out, and any odd number of negative symbols will cancel
out all but one negative.
Primary Rules of Exponents
1. If a > 0, then a > 0.
2. If a < 0 and n is odd, then a < 0.
3. If a < 0 and n is even, then a > 0.
4. 0 = 0
5. 1 = 1
6. a = 1
7. a = a
8. a b = (ab)
9.
= ; b ≠ 0
10. a a = a
11.
= a ; a ≠ 0
12. (a ) = a
Rule 2 can be illustrated with (−2) = (−2)(−2)(−2) = −(2 ∙ 2 ∙ 2) = −(2 ) = −8 because two of the
negative symbols cancel each other out. On the other hand Rule 3 can be illustrated with
(−2) = (−2)(−2)(−2)(−2) = 2 ∙ 2 ∙ 2 ∙ 2 = 2 = 16. As long as the base or argument is not zero, a
good rule of thumb to remember is that even exponents always yield a positive result, and odd
exponents always yield a result with the same sign as the argument.
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Example 1 (Proof of Rule 8)
Prove: a b = (ab) .
Proof: a b = aaabbb = ababab = (ab) . Q.E.D.
Note: The acronym Q.E.D. stands for the Latin phrase Quod Erat Demonstratum, which roughly
translates to “That which has been proven.” Often, a box, ∎, will be written to represent Q.E.D.
Example 2 (Proof of Rule 9)
Prove:
= .
Proof:
= = = . Q.E.D.
Example 3 (Proof of Rule 10)
Prove: a a = a .
Proof: a a = (aaa)(aa) = aaaaa = a . Q.E.D.
We will save the proof for rule 6 until after we delve a little deeper into exponents.
Example 4 (Proof of Rule 12)
Prove: (a ) = a .
Proof: (a ) = a a a = (aa)(aa)(aa) = aaaaaa = a . Q.E.D.
You may notice, we specify on rules 9 and 11 that the denominator cannot equal 0. This is due to the
First Commandment of Mathematics.
1 st COMMANDMENT: THOU SHALT NOT DIVIDE BY 0. WHEN THOU DOST, THOU SHALT
BLOWETH UP THE EXPRESSION. (Technically we consider it undefined, but explosions are nifty)
Exponential Arithmetic
You can only add two exponentials together if they have the same base and the same exponent.
ac + bc = (a + b)c
You can only multiply two bases together if they have the same exponent. (Rule 8)
a b
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Example 5 (Adding with exponents)
3x + 4x − 5x + 6x = (4 + 6)x + (3 − 5)x = 10x − 2x
Example 6 (Multiplying with exponents)
5(3) ∙ 2(6) = (5 ∙ 2)(3 ∙ 6) = 10 ∙ 18
Radicals
Radicals are also known as roots to most people. Their creation was to cancel out or “undo”
exponentiation. The most common radical is the square root, symbolized by √, which was intended
to undo squares. For every exponent, there is a corresponding root, but rarely do you see many of
them before Algebra 2. For instance the inverse operation of cubing, a
, is cube rooting, √a. The
number inside the radical being “undone” is called the radicand, but for our purposes, we shall refer
to it as an argument.
Because even exponents yield positive results, even radicals can only “undo” positive numbers, and
since odd exponents keep the same sign, so do odd radicals. This results in the second
commandment of mathematics.
2 nd COMMANDMENT: THE ARGUMENT OF AN EVEN ROOT SHALT ALWAYS BE GREATER
THAN OR EQUAL UNTO ZERO.
Taking an even root of a negative number results in an imaginary number, a subset of all numbers
that is a result of mathematicians getting bored with real numbers applicable to the real life they don’t
possess. But not to worry, the next commandment will not show up until much later.
The key to radicals is to understand that they are fractional exponents as evidenced by the following
rule:
We can expand this to show two more rules:
13. √x
14. √x
= √x = x
= √x = x
Example 7 (Proof of Rule 13)
√a
= a
Prove: √a
Proof: √a
= √a = a .
= (a ) = a
and √a = a
= a .
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Simplifying Radicals
What if we are given the problem, √50 + √8? We would probably reach for our calculators in order to
get a decimal approximation. However, there is an acceptable simplification that can be performed
on this problem. If we remember that square roots undo squares, then all we have to do is look for
square factors of the argument.
Example 8 (Adding Radicals)
√a b
= a √b
√50 + √8 = √25 ∙ 2 + √4 ∙ 2 = 5 ∙ 2 + 2 ∙ 2 = 5√2 + 2√2 = 7√2
Important note: Once we get both terms in terms of √2, then we can follow the guidelines of adding
two exponents together: i.e. Same base (2) and same exponent .
Example 9 (Multiplying Radicals)
5√6
3√24√7 = (5 ∙ 3 ∙ 4)√6 ∙ 7√2 = 60√42√2
The final answer in this case might not look pretty, but it is simplified fully.
Rationalizing the Denominator
Myth: Radicals in the denominator are evil.
Fact: Radicals in the denominator are perfectly acceptable.
However, most multiple choice tests involving radicals rationalize the denominator. So, it is fairly
important that we understand what that means. Conceptually, we would like to cancel out the radical
in the denominator.
Historically, this has been taught as follows:
We would like to rationalize
√
. We do so by multiplying by a clever form of 1, namely . By doing
so, we get
√ ∙ √
√ = √
√
= √
√
. Essentially, we needed to cancel out the square root by squaring it.
Things get a little hairier when we have to rationalize cube roots and fourth roots. We can then use
this general rule:
1
√a
= √a
a
√
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Example 10
Rationalize
√ .
10
√5
= 10 √5
5 = 2√5
Example 11
Rationalize
10
√5
.
√
= 10 √5
5 = 2√25
Negative Exponents
When the exponent itself is negative, this has nothing to do with the sign of the answer. A negative
exponent is considered a reciprocal operation.
a = 1 a
Example 12
3
5 = 5
3
Example 13
a b
c
= a c
b
= 5
3 = 25
9
Think of the fraction bar as a fence, and remember the old adage “The grass is always greener on the
other side of the fence.” An exponent is negative until it crosses to the other side of the fence. Then
it becomes positive.
Example 14 (Proof of Rule 11)
Prove:
= x .
Proof:
= x
= x x = x
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Example 15 (Proof of Rule 9)
Prove: =
.
Proof: = (ab ) = a (b ) = a b =
.
Cancelling Out
Cancelling can only occur if you have identical factors in both the numerator and denominator.
It is a common misconception that the x in
can be cancelled out leaving -5. This is false, as x
is not a factor in the numerator. Now if we had ()
, then the x can be cancelled, since the x is
now a factor in both the numerator and denominator.
Example 16a
a b
a b = a b
a b = b
a
In this case, the b’s need to be moved first. Then two of the a’s cancel out leaving three of them in
the denominator. Three of the b’s cancel out leaving four of them in the numerator.
Example 16b
a b a
a =
b a ∙ b
b = a b () = a b = b
a
Example 17 (Proof of Rule 6)
Prove: a = 1.
Proof: a = a =
= 1.
A student reading this text may be wondering if every section in this book will have as many
examples and explanations as the last three sections have had. The short answer is no. The more
lengthy explanation would make this section even longer than it already is, so it will be left to our
imagination. If one’s imagination should evoke images of blue skies and white puffy clouds, this has
nothing to do with the lengthy explanation we are trying to avoid. One may point out the fact that the
lengthy explanation we are attempting to avoid here may have taken less space than the explanation
of its avoidance. To said person, we say “Go daydream about blue skies and white puffy clouds.”
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I.A.3 Independent Practice
Simplify the following by eliminating the parentheses and combining all like terms.
1. 5a + 3a + 2b − 4a
2. 4c − 2c + 3c
3. r + r + 3r − 6r
4. 6d + 5d − 2d
5. 4x − 2x + 5
6. 2t + 3s − 5t + 2s
7. 3√2 + √8
8. 4√3 + √45
9. 3√20 − 2√45
10. √16
11. √81
+ √24
− √24
12. 4√20 + 6√12
13. (3y )(5y )
14. (2xy )(6x )
15. (5a b )(8ab )
16. 7√53√6
17. 3√66√6
18. 2√4(5√6)
19. (r )
20. (3a )
21. (5v )
22. (3p )
23. (2a b )
24. (3g d f)
25.
26.
27.
28.
29.
30.
Rationalize the following.
31.
√
32.
√
33.
√
34.
√
35. √
√
36.
√
37.
√
38.
√
39.
√
40.
√
41.
√
42.
√
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Section
I.A.4
Order of Operations
(GEMA)
The concept of Order of Operations has been taught for generations. It is described as the agreedupon
order in which an expression must be simplified. Up until now, it has always been defined by
the acronym PEMDAS, which stands for Parentheses Exponents Multiplication Division
Addition Subtraction. A mnemonic device has been used to help remember the order of
operations, “Please Excuse My Dear Aunt Sally”. There are several flaws in how this is commonly
taught.
No one is quite sure what Aunt Sally did, but suffice to see, with her notoriety in many
mathematics courses, it was bad.
Addition and Subtraction actually happen in the same step from left to right. Likewise,
Multiplication and Division occur in the same step as well from left to right. However, this is
commonly forgotten, and addition is performed before subtraction, and multiplication before
division. For instance 5 − 2 + 3 = 6. If addition occurs before subtraction, the answer is 0,
which is, in fact, wrong.
There is no allowance for radicals in PEMDAS.
Parentheses aren’t always written when a group of operations must occur first.
It’s at this point we modify the order of operations to GEMA, which now stands for Groupings
Exponents Multiplication Addition. You could even create a mnemonic device for this like
“Gorillas Eat Moldy Apples”. Or you can just remember to go from complex to easy, working from
inner most groups outward.
Groupings
Groupings are not just parentheses. Groupings constitute a set of operations that may be rewritten
with parentheses to indicate their grouping. In fact, when inputting an operation into a calculator, the
optional parentheses must be used, so it is good practice to rewrite the problem with parentheses.
Groupings can be set apart by any of the following:
Parentheses – Every open parenthesis needs a closing parenthesis. ( )
Brackets – Every open bracket needs a closing bracket. [ ]
Fraction Bars – The numerator is one group, and the denominator is another. In a calculator,
the division key is used for the fraction bar. We must put both the numerator and denominator
in their individual sets of parentheses.
Radicals – Anything inside of a radical is considered grouped within the radical. The radical
button on the calculator always provides an open parenthesis for our convenience.
Exponents
Now that we know exponents and radicals are in fact the same thing, we can perform them in
whatever order we feel like, after converting all radicals to exponents of course.
Example 1
√4 = 4
= 2 = 8 or √4 = 4
= (4 ) = 64 = 8
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Multiplication
We have also learned that division can be rewritten as multiplication of a reciprocal. Therefore a
problem like 5 ÷ 2 × 6 = 5 × × 6 = 5 × 3 = 15. As is evident, once everything has been simplified to
multiplication, the order in which we multiply does not matter. And in this case, it is much easier to
multiply the by 6 first rather than deal with a fraction like , which we would have to do if we
maintained the old Order of Operations.
Addition
Addition is similar to both multiplication and exponentiation in the fact that subtraction can be
converted to addition by adding a negative number. For example:
2 − 5 + 8 = 2 + (−5) + 8 = 10 + (−5) = 5. Once all subtraction has been eliminated, we can add the
result in any order. In this case, for some people, subtracting 5 from a smaller number is
conceptually difficult, but if we add our two positive values first, it seems to help some people out.
Putting it all Together
Essentially, we need to convert everything into exponents, multiplication, and addition. If we prefer,
we can place parentheses in appropriate locations. Then we simplify all groups, then work from
complex (exponents) to simple (addition). From now on, we will refer to individual values being
added together as terms. We have seen instances where we cannot add terms together, and that is
okay.
Example 2
3 + 5
5 − 2 + 3√4 = (3 + 5)
5 + (−2) + 3(4)
Example 3
(9 + 5)
= + 3(4)
3
= 14
3 + 3(4)
= 14
3 + 3(2)
= 14
3 + 6
= 32
3
3 + 5 + 3(2) = √28 + 3(2) = 2√7 + 6
This is fully simplified. There’s technically no reason to simplify √28 since it cannot be simplified to
an integer, and there is no other √7 in the problem.
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I.A.4 Independent Practice
Simplify the following as fully as possible.
1. 5 + 2(4) − 3 =
2. 3 − √9 + 5(2) =
3.
() =
4. 5(3) + 7
+ 5 =
5. √4 + 4 + 2(5 )
=
6. (6 + 3) − 8(4) =
7. (3 − 8) +
() =
8.
√ =
9. a + √a =
10. 3a + 15√a
=
11. √
=
12. 15x + 7(x ) − 8x + (2x ) =
13. 6 7 a4(3 + 2 ) =
14. √7 − 3 =
15. 45 + 2(3 + 5) =
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Section
I.A.5
Solving Multi-Step
Equations and Inequalities
Many students find solving equations confusing and difficult, mainly due to the fact that the Order of
Operations has been made into a much more complicated process than is absolutely necessary.
There are two different possibilities for most equations. Either they have variables on only one side of
the equation or they have variables on both sides of the equation.
One-Sided Equations
Simplify both sides of the equation separately using GEMA.
Apply the reverse order of operations, AMEG, working outside inward to undo the equation
and get x by itself.
Anything you do to the side with x, you must do the other side of the equation to keep
everything equal.
If “undoing” an even exponent, you must apply a ± to stand for “positive or negative.” Since both
5 = 25 and (−5) = 25, then when solving for the square root of 25, we could get either 5 or −5.
However if √25 is part of the actual equation, we assume the answer is simply 5.
Example 1
Solve (5 + 3)x − 5 = −2.
(8x + (−5)) = −2
8x + (−5) = (−2)
8x = 4 + 5
x = 9 8
x = ± 9 8
x = ± √9
√8
x = ± 3
2√2
x = ± 3√2
4
Technically speaking, ± is a perfectly acceptable place to stop.
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Distribution
There are times when the only way to simplify a problem is through a process called distribution.
The process is written as follows.
a(b + c) = ab + ac
In English, we take a and multiply it by every separate term inside the parentheses. Very often in
simplifying we must distribute left to right as written above. There are occasions, when we must
“reverse-distribute” from right to left in order to solve a problem.
WARNING: a(b + c) ≠ (ab + ac) . Remember, we cannot multiply two bases together unless their
exponents are the same. In this case the exponent of a is 1 and the exponent of (b + c) is 2. If we
had a rather than just a, we would end up with a (b + c) = (ab + ac) , however this is not as
straightforward as the original formula. It is complicated by the fact that a (b + c) = a b + a c is a
perfectly acceptable equation because when the sum’s exponent is 1, we can distribute anything we
like into it.
Example 2
Solve 5x + x√3 = 6 for x.
We cannot combine the two x-terms in this problem because one of them has a square root of 3
attached, but we can recognize that both terms have an x, and therefore we can reverse-distribute it.
5x + x√3 = 6
x5 + √3 = 6
x = 6
5 + √3
This is a perfectly acceptable answer, since 5 + √3 has a numerical value. It may not look pretty, but
we will leave our answer as is. Some mathematicians may cry about the radical in the denominator,
but for right now, we will let them.
Dual Sided Equations
Simplify both sides of the equation.
Multiply by a common denominator to get rid of any fractions.
Move all x terms to one side of the equation and all non-x terms to the other side.
Now treat the equation like a one-sided equation.
For right now, all of our equations will be solvable using AMEG. There are some equations that
require higher level algebra, and there are some equations that cannot be solved algebraically. For
the most part, the equations you will see on standardized tests and in the real world can be solved
with the steps set forth here.
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Example 3
Solve 5x + 2 = 43 + x√2 for x.
5x + 2 = 12 + 4x√2
5x − 4x√2 = 12 − 2
x5 − 4√2 = 10
x = 10
5 − 4√2
Example 5
Solve x + = x − 4 for x.
A common multiple of 5 and 2 is 10. We then distribute 10 across all terms in the equation.
x + 1 5 = 3 2 x − 4
10 x + 1 5 = 10 3 x − 4
2
10x + 2 = 15x − 40
42 = 5x
x = 42
5
Inequalities
The symbol, 3, because 3 is smaller than 5 and thus must get the smaller end
of the symbol.
Things get trickier when variables are introduced. x = 5 implies that both x and 5 are the same.
However, x < 5 implies that x is any value smaller than 5. This means that any value we can think of,
as long as it is less than 5 is a possible solution to this problem. You will often see this graphed as:
We call 5 an exclusive endpoint due to the fact that 5 is excluded from the answer. If we wish to
include 5 as part of the answer, we use the symbols ≤ and ≥, which imply inclusion. In other words,
the two sides can be but are not necessarily equal.
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Solving Inequalities
Solve an inequality as if it were an equation. The only subtle difference is that if it becomes
necessary to multiply or divide by a negative value, the direction of the inequality symbol must be
reversed. Why? We know 3 < 5, but if we multiply both sides by -1, we end up with −3 > −5.
Notice, the symbol must switch, or we get a false statement.
For right now, we probably will not see too many exponents in our inequalities. Exponents cause all
sorts of problems that make solving inequalities more complicated than we are ready for now. We
will come back to this when we discuss quadratic inequalities and absolute value.
Example 4
Solve the inequality 6x − 5 > 10x + 4 for x.
6x − 5 > 10x + 4
6x − 10x > 4 + 5
−4x > 9
x < − 9 4
Notice that the inequality switched directions because we divided by -4.
There will be some problems that appear to be solvable using AMEG. Many students attempt these
problems using very clear logic, but they forget one thing. Here is such an example.
Example 5 (This Example is Intentionally Wrong)
Solve 10x = 80√x for x.
10x = 80√x
10x = 80x
10x = 80
10x = 80
x = 8
x = 8
x = 4
This is a correct answer. However it is only one of two correct answers. The second one is missing
because we divided out x . Since there is a possibility in our original problem that x could equal 0,
we cannot willy nilly divide out x’s. We may very well have divided by 0, which violates the first
commandment of mathematics. As a general rule, we cannot divide by x unless we are told explicitly
that x cannot equal zero. The proper way to solve this problem will be investigated further in chapter
I.C.
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Absolute Value
Before we solve an equation involving absolute value, it is important that we understand what it
means. The absolute value of a number is its distance from 0. Since distance is always a positive
value, the absolute value is always positive. The notation for absolute value is two vertical bars that
act as a form of grouping. For instance, |−3| = 3 and |5| = 5. When solving an equation or
inequality involving absolute value, things can get a little tricky.
Example 6
Solve |3x + 2| = 11 for x.
Since |3x + 2| = 11, we get two different equations and thus two different solutions.
3x + 2 = 11 3x + 2 = −11
x = 3
x = −
Example 7
Solve |2x − 1| ≤ 5 for all values of x.
Again, this is actually two different inequalities.
2x − 1 ≤ 5 2x − 1 ≥ −5
x ≤ 3
x ≥ −2
Notice, that when the 5 changed signs, we also changed the inequality.
Example 8
Solve 5(x + 2) − 3 > 42 for all values of x.
This one may not look like it involves absolute value, but looks can be deceiving. First, let’s add 3 to
both sides and then divide both sides by 5. This will result in (x + 2) > 9. In the next step we will
square root both sides, but it’s tricky when mixed with inequalities. Square rooting both sides we get
|x + 2| > 3, which results in
x + 2 > 3 x + 2 < −3
x > 1
x < −5
The good news is this is as complicated as this gets. We will not see an equation with an absolute
value on both sides of the equation, at least not in this course. If we are presented this type of
problem in future mathematics courses, a graphing calculator should help quite a bit.
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I.A.5 Independent Practice
Solve the following equations for the variable.
1. 5x − 3 = 8
13.
=
2. √x + 5 + 3 = 11
14. 3(x + 5) = 2(x − 6)
3. x(5 + 4 − 2) = 38
15. 5(x − 2) ≤ √2(x + 5)
4. 6(x − 2) = 24
16. 7x + 4 < 8x − 6
5. 5x − 2 = 13
17. 6x + 2 > 2x − 8
6. 6(2) + x − 5 = −23
18. x + 5 ≥
7. (3 + 5x) − 3 = 5
19.
≤
8. x√36 − 5√2 = 7√3
20. 6x − 5 < x3 + √2
9. 4x + 5 = 7
21. |8x + 2| < 18
10. 7√2x + 2 − 11 = 3
22. |3x − 5| ≥ 2x + 4
11. |5x + 6| = 3x − 7
23. 3x + 5 ≤ 53
12. + 3x = x − 2 24. (3x + 5) > 121
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Cogswell – PCM

Chapter
I.B
39

Chapter
I.B Chapter
Overview
‣ I.B.1 – Function Overview
This section investigates the idea of functions conceptually and applicably.
‣ I.B.2 – Graphing a Linear Function
This section describes how to graph a line from linear equation in slope-intercept form, standard
form, and point-slope form.
‣ I.B.3 – x-intercepts and y-intercepts
This section shows how to find the x-intercepts and y-intercepts given linear equations in slopeintercept
form, standard form, and point-slope form.
‣ I.B.4 – Finding an equation given a slope and a point
or two points
This section eliminates the need for using the slope formula and provides a quick mental-math
alternative for finding the equation of a line.
‣ I.B.5 – Parallel and Perpendicular
This section investigates the ideas of parallel and perpendicular lines in relation to their equations
in slope-intercept form, standard form, and point-slope form.
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Section
I.B.1 Function
Overview
Function is yet another dirty mathematical F-word. This is usually due to a less-than-firm foundation
in the concept of functions. A function is a device made up of various components that transform an
input into a corresponding output. A generator is a device made up of smaller parts that takes in one
type of energy (mechanical or chemical) and creates electrical energy. An assembly line in a factory
made up of people and/or robotic components will take in materials and transform them into a product
for sale. A dog made of biological components takes in food and converts it into… well, you know…
Functions are clearly defined so that every input has one specific corresponding output. There are no
surprises. People would not purchase a generator if there is a chance it will produce fire instead of
electricity. A car factory will go out of business if every once in a while it creates a pie. A dog would
be considered unusual if it laid gold nuggets every so often in the backyard. Granted, it would make
paying for the upcoming vet bill more manageable.
In mathematics, functions are defined by a set of arithmetic operations. Traditionally, x is considered
the input, and y is considered the output. The function y = 3x + 5 takes in an x-value, squares it,
triples it, and then increases it by 5 to produce a y-value. The y-value is never random. It is strictly
determined by the x-value plugged into the equation. In mathematics, we say that the y-value is
dependent on the x-value. Because the x-value can be any value we choose (as long as it does not
violate the Commandments), we say that it is independent. The set of all x-values that do not violate
the Commandments are considered the Domain, and the dependent y-values are considered the
Range.
Example 1
What is the domain of y =
?
We know that we cannot divide by 0 (Commandment 1). Therefore x − 3 ≠ 0. Solving this we find
that x ≠ 3. Therefore the domain is all real numbers except for 3. Either we write D: {R|x ≠ 3} or
D: {(−∞, 3) ∪ (3, ∞)}. “D: {R|x ≠ 3}” means that the domain (D) is all real numbers (R) when x is not
equal to 3 (x ≠ 3). D: {(−∞, 3) ∪ (3, ∞)} is interval notation and means all values except 3. In
interval notation parentheses indicate exclusion, while brackets indicate inclusion.
Example 2
What is the domain of y = √x + 5?
We know that we cannot take an even root (in this case square root) of a negative value. Therefore
x + 5 ≥ 0 or x ≥ −5. The domain is all real numbers greater than or equal to -5. We can write this as
D: {R|x ≥ 5} or D: {[−5, ∞)}. Notice in this set of interval notation, we are using a bracket to include
−5 as one possible solution.
Note: ∞ and −∞ are never included and, thus, will never be written with a bracket. The reason for
this should be obvious since one can never actually reach and, therefore, never equal infinity.
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Points
A point is a position defined by a set of coordinates. Traditionally, functions are restricted to the x-y
plane, defined by a horizontal axis, x-axis, and vertical axis, y-axis, intersecting at a point called the
origin, whose coordinates are (0,0). A coordinate is made up of an x-coordinate and a y-coordinate.
These are set apart by a comma and grouped inside a set of parentheses with the x-coordinate
always written first. Be careful not to confuse points with interval notation.
To the left of the origin all points have negative x-coordinates, and to the right of the origin all points
have positive x-coordinates. Below the origin all points have negative y-coordinates, and above the
origin all points have positive y-coordinates.
Example 3
Where is the point (3,-2) located?
The point is located 3 units to the right and 2 units below the origin.
Example 4
Where is the point (0,5) located?
The point is located 5 units above the origin. Since the x-coordinate is 0, there is no horizontal
translation from the origin.
Graphs
A graph is a set of points defined by an equation. For an unknown equation, it is easiest to
determine points that satisfy the equation and connect them.
Example 5
Graph the equation y = 3x − 2.
Typically, we would want to find as many points as possible, but in the interest of time we will find
only a few points. To accomplish this, we will substitute several values for x into the equation and
find their corresponding y-values. Once this is accomplished, we can then plot the points and create
a possible graph. To get started, we will substitute -2 in for x. 3(−2) − 2 = 3(4) − 2 = 12 − 2 = 10.
Other corresponding x and y-values are written in the table below.
y
y
X -2 -1 0 1 2
Y 10 1 -2 1 10
x
x
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Remember, a function is a relation in which every input has one and only one corresponding output.
In other words, each x-value has only one corresponding y-value. If a graph has at least two points
with the same x-coordinate, it is not a graph of a function.
Example 6
y
Is this a graph of a function?
x
It is a function since there are no two points with the same x-coordinate.
Example 7
y
Is this a graph of a function?
x
This is not a function. There are multiple points that share x-coordinates.
Note: The vertical line test is an excellent method for determining a function. All points on a vertical
line share the same x-value. So if a vertical line can be drawn anywhere on a graph so that it
intersects the graph at more than one point, then the graph is not a function. Likewise, if every
vertical line that can be drawn intersects the graph once only, the graph is a function.
Function Notation
Function Notation is a useful tool, but many students do not truly appreciate its usefulness. If there
are multiple functions all of the form “y =”, how do we distinguish between them. The only way we
possibly could is by rewriting their equation while describing them. For instance, if there are two
equations y = 3x + 2 and y = 5x − 1, and we are asked to plug 3 in for x, we would ask “Which
equation?” We would have to be told “y = 3x + 2” or “y = 5x − 1”. This seems like a waste of time.
We could always assign names to these equations. We will let y = 3x + 2 be f, and y = 5x − 1 be g.
Now when we ask “Which equation?” the response will be either f or g. This eliminates some of the
wasted time, but there is still the need to ask “Which equation?” We could always be told “Substitute
3 for x in the f equation”, but there has to be an easier and more mathematically confusing way to
write this. Well, mathematicians came up with a way.
Given the functions f(x) = 3x + 2 and g(x) = 5x − 1, we can now distinguish between the two
equations. An interesting side effect is that in writing the function, we can now indicate the input.
Breaking it down, f is the name of the function, x is the input, f(x) is the output, and 3x + 2 is the
definition of f. Perhaps we still cannot truly appreciate this until we see an example or two.
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Example 8
If f(x) = 5x − 2, determine f(2).
What are we being asked? Determine the output that results from inputting 2 into the f function. In
other words f(2) = 5(2) − 2 = 5(8) − 2 = 40 − 2 = 38. Therefore f(2) = 38. If we were to graph f,
we know (2,38) would have to be a point on the graph.
Example 9
Graph the line g(x) = 3x − 2, by finding g(0) and g(2) and connecting the two points.
g(0) = 3(0) − 2 = 0 − 2 = −2 and g(2) = 3(2) − 2 = 6 − 2 = 4. Thus we have the points (0, −2) and
(2, −4). We can now graph these two points and the line that connects them.
y
x
Calculator Interlude
The Y= screen is the most powerful thing on the calculator. You can input an equation, and from that
equation the calculator can graph it, provide a table of values, or provide individual values for use in
formulas.
After an equation is inputted into Y1, we can go to TBLSET, which allows us to set up the table
however we want. TblStart is the x-value we want to start with. ∆Tbl is the interval by which we
want to count. If we want to input even positive numbers, we would set TblStart = 2 and ΔTbl = 2.
Then we can go to TABLE and see the results.
Another option is to go to VARS Y-VARS 1: Function Y1 to actually use Y1 as function
notation.
Example 10
Given f(x) = 3x − 8, use a calculator to determine f(5).
In Y= we set Y1=3x^5-8. Then we press QUIT. Next we go to VARS Y-VARS 1: Function
Y1, and on the home screen input Y1(5) and press ENTER. We find that f(5) = 9367. We could
also look up the answer in the table.
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I.B.1 Independent Practice
Determine the domains of the following functions.
1. f(x) =
2. f(x) = √x − 4
3. f(x) =
4. f(x) = √x + 5
5. f(x) = x − 9
6. f(x) =
√
Given f(x) = 3x and g(x) = 2x + 10, determine the following.
7. f(2) =
8. g(−3) =
9. f(−5) =
10. g =
11. f(3x) =
12. g(x + 2) =
Fill in the tables and graph the following.
13. f(x) = 5x − 4
x
-2
-1
0
1
2
y
y
x
14. g(x) = x − 3
x
-4
-2
0
2
4
y
y
x
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Section
I.B.2
Graphing a Linear
Function or Inequality
A linear function is the simplest function that takes in an x-value and produces a y-value. Even
though it has extremely important applications, it is, in and of itself, extremely simple. It is a wonder
that so many math courses spend so much time with these functions until you consider the fact that
most questions on standardized tests, like the SAT and TAKS, utilize linear equations. The process
of teaching linear functions has gotten so lengthy and complex that most students do not understand
the basic concepts, and less time is available to learn about more complex topics.
Slope
Possibly, the most important feature of a line is its slope. The slope is so important, in fact, that half
of Calculus studies slopes and how to calculate them. In calculus, we refer to these slopes as
derivatives. In application, a slope is a rate of change like miles per hour, meters per second, or
revolutions per minute. Notice, the word per still means divide, so that rates of change like miles per
hour literally mean “miles divided by hours”.
A line is unique in the fact that its slope, as well as its rate of change, is constant. The slope of a line
is commonly referred to as m, but it does not matter what letter we use to represent slope, because
the majority of the time, we will attempt to calculate an exact numerical value. Slope is defined by
any of the following:
Slope = rise
run = y − y
= y − y
= Δy
x − x x − x Δx
The formulas are useful but not necessary. In some cases, they can actually be confusing. What we
really need to remember is as we go from left to right, the y-values increase, then the slope is
considered positive. If the y-values decrease from left to right, then the slope is negative.
Example 1a (Formulaic)
Find the slope between the points (−2,6) and (4, −2).
Using the first formula:
= ()
=
= − .
Using the second formula:
=
= − = − —
The formulas work, but there is a lot of room to make errors with all of the negative signs.
Example 1b (Logical)
Find the slope between the points (−2,6) and (4, −2).
We can see that as we go from left (-2) to right (4), the y-values decrease. Therefore, the slope is
negative. The distance between 6 and -2 is 8, and the distance between -2 and 4 is 6. Therefore our
rise is 8 and our run is 6. The slope is − or − .
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If we consider the logic behind the concept of slope, our answer will never be wrong.
Example 2
Find the slope between (3,-4) and (-5,-7).
Going from left (-5) to right (3), our y-values increase. Therefore the slope is positive. Now the
distance between 3 and -5 is 8, and the distance between -4 and -7 is 3. We, thus know that the
slope is .
Example 3
Find the slope between (4,7) and (8,7).
Going from left (4) to right (8), we see that the y-values do not change. Therefore the rise is 0 and
the run is 4. The slope is = 0. All horizontal lines have a slope of 0.
Example 4
Find the slope between (5,3) and (5,-9).
We cannot go from left to right because the x-values do not change. So the rise is 12 and the run is
0, which would mean the slope is . This violates Commandment 1, so the slope in this case is
undefined. All vertical lines have an undefined slope.
Slope-Intercept Form
The slope-intercept form of a linear equation is widely accepted as y = mx + b or y = ax + b. The
easiest way to graph a line in this form is to choose to x-values, find their corresponding y-values, plot
the points, and draw the line between them.
Example 5
Graph f(x) = x − 2.
We need to choose two x-values. Because the coefficient of x is , it
would be smart to choose to x-values that are divisible by 3. We can
choose 0 and 3. f(0) = (0) − 2 = −2 and
f(3) = (3) − 2 = 0. Plotting (0,-2) and (3,0), we can then graph the line
between them.
y
x
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Linear Inequalities
Linear inequalities graph in the same manner as linear equations with a couple of twists. For one,
when y cannot equal the given expression (i.e. < or >), the line is dotted or dashed. Additionally, to
graph all values that satisfy the inequality, we must shade the solution area.
Example 6
Graph y < x − 2.
Plugging in 0 and 3 in for x, we get “points” (0, −2) & (3,0). Since y is not
equal to the expression we graph a dotted line through these two points.
Additionally, since y is less than the points in the dotted line, we shade
below the line.
Note: The use of quotation marks around the word points is necessary because points indicate values
where y is equal to the expression. Although these locations aide in graphing the line, they are not
true points due to the lack of equality.
Example 7
Graph y ≥ − x + 1.
Plugging in 0 and 2 in for x, we get points (0,1) & (2, −2). The line will be
solid since y can equal the expression, and the shading will be above the
line since all y-values in the solution are bigger than the line.
Standard Form
The standard form of a linear equation is of the form Ax + By = C where A, B, and C are all integers.
There are many textbooks and classrooms that encourage students to rewrite a standard form
equation into slope-intercept form, but this is mostly unnecessary. One method is to substitute 0 for x
and solve for y and then substitute 0 for y and solve for x. This will provide two points that we can
use to graph a line. Technically, we can substitute any values in, but zero typically is the easiest
option.
Example 8a (Converting first)
Graph 3x + 2y = 9.
If we solve this for y in terms of x, we get:
3x + 2y = 9
2y = −3x + 9
y = − x +
Substituting 0 and 2 for x we get the points (0,4.5) and (2,1.5), which we
can use to graph the line.
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y
x
Example 8b (Alternate Method)
y
Graph 3x + 2y = 9.
Substituting 0 in for x, we get 2y = 9 or y = 4.5. This is the point (0,4.5).
Substituting 0 in for y, we get 3x = 9 or x = 3. This is the point (3,0).
x
Example 9
Graph 2x − 5y > 10.
The intercepts are (0, −2) & (5,0). We connect the two with a
dotted line (No ‘=’). Here is where the tricky part comes in. Do we
shade up or down? Plug in a value not on the line; any point will do.
If plugging the point in results in a true statement, we shade toward
the point. Otherwise, we shade away from it. So if we plug in the
point (0,0), we end up with the statement 0 > 10, which is false. So
we shade away from the point. In other words, we shade down.
Horizontal and Vertical Lines
All points on a horizontal line have the same y-value (See Example 3). Therefore the equation for a
horizontal line is y = a or f(x) = a, where a is the constant y-value that all points on the horizontal
line share. For this reason f(x) = a is commonly referred to as a constant function.
Likewise, all points on a vertical line have the same x-value (See Example 4). Therefore the equation
for a vertical line is x = a. Since a vertical line fails the vertical line test (There are an infinite number
of points that share the same x-value), a vertical line is not a function and is often referred to as a
constant equation.
Example 9
Graph x = 2.
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I.B.2 Independent Practice
Find the slope of the line connecting the pairs of points.
1. {(3,5), (6, −3)}
2. {(2,4), (7,11)}
3. {(5,4), (−2,8)}
4. {(−1,5), (3,9)}
5. {(3,7), (3,2)}
6. {(5,2), (9,2)}
Graph the following on the coordinate planes provided.
7. y = 3x + 4
y
7. 8.
y
8. y = x − 3
x
x
9. y = −4x + 5
10. y = − x + 7
11. 2x + 5y = 15
y
9. 10.
y
12. 3x + 4y = 24
x
x
13. 5x − 8y = 40
14. 2x − 7y = 14
y
11.
12.
y
x
x
y
13.
14.
y
x
x
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Section
I.B.3
x-intercepts and
y-intercepts
Technically, we have already found x-intercepts and y-intercepts when we substituted zero in for x
and y in the last section. y-intercepts are extremely easy to find, but are not that useful in the grand
scheme of things, whereas x-intercepts are slightly more difficult to find and are the basis of most
graphical algebra. With linear functions, both are fairly simple to find, but the overall idea is
consistent with the more complex functions we will see later on.
y-intercepts
A y-intercept is the point where the graph crosses the y-axis. Every function should only ever have
one. If a graph has more than one, it is not a function as it fails the vertical line test (The y-axis being
the vertical line). The y-axis (and by extension all vertical lines of the form x = a) has an infinite
number of y-intercepts, as every point on the y-axis “intersects” the y-axis. Every point on the y-axis
is of the form (0, b), where b is a constant. This essentially means that the x-value for every point on
the y-axis is 0. Therefore, to find a y-intercept, we substitute 0 for x and solve for y.
Example 1
Find the y-intercept of f(x) = 5x + 2.
f(0) = 5(0) + 2 = 0 + 2 = 2. The y-intercept is 2.
Note: The reason y = ax + b is referred to as the slope-intercept form is because a is the slope and b
is the y-intercept.
x-intercepts
An x-intercept is the point where the graph crosses the x-axis. The x-intercept is also known as the
solution to the graph or root of the graph which occurs when the y-value is set equal to 0, which
incidentally is how we solve it. Many advanced functions have multiple x-intercepts, but linear
functions only have one. The exception to this rule is horizontal lines. The x-axis (y = 0) has an
infinite number of x-intercepts while all other horizontal lines have no x-intercepts. Graphically, a
horizontal line is parallel to the x-axis and so, by definition of parallel, does not intersect the x-axis.
Algebraically, we cannot set a constant other than 0 equal to 0. The universe would implode in on
itself.
Example 2
Find the x-intercept of f(x) = 3x + 7.
0 = 3x + 7 → −7 = 3x → x = − .
Finding the x-intercepts might seem less than challenging now, but it is good to understand their
importance now before we graduate to functions with multiple solutions.
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I.B.3 Independent Practice
Calculate the x-intercept, slope, and y-intercept of the following.
1. f(x) = 3x + 2
2. g(x) = 4x − 5
3. h(x) = x − 6
4. f(x) = − x + 2
5. g(x) = x − 6
6. h(x) = − x + 2
7. 3x + 5y = 15
8. 4x + 7y = 28
9. 2x + 6y = 6
10. 5x + 3y = 12
11. 2x − 7y = 15
12. 5x − 8y = 10
13. 4x − 2y = 6
14. x − 3y = 7
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Section
I.B.4
Finding an Equation Given a
Point and Slope or Two Points
One of the most important skills a student needs to master is the ability to create an equation from a
set of discrete points. This process is known as regression. For a linear equation we only need two
points, since any two points are collinear. If we are given a single point, then we need to know the
rate of change (slope) as well. There are several techniques that are commonly used for this
process: a couple pencil and paper techniques and a mental one.
Example 1a (Point-Slope Method)
Determine a linear equation with a slope of that contains the point (10,2).
For this method we will use the point-slope form of a line, namely: y − y = m(x − x ). Therefore we
get y − 2 = (x − 10). Distributing, we get y − 2 = x − 4, and adding 2 to both sides, we get
y = x − 2.
Pros: We simply have to plug in the given values and solve for y.
Cons: We have to memorize yet another form of a linear equation. Also, we have to make sure we
plug the point values into x and y only and not into x and y.
Example 1b (Slope-Intercept Method)
Determine a linear equation with a slope of that contains the point (8,-3).
For this method we will use the slope-intercept form of a line. We know that the line will be of the
form y = x + b where b is the y-intercept. Plugging in the values x = 8 and y = −3 we get
−3 = (8) + b. This simplifies to −3 = 12 + b, and solving for b, we get b = −15. Therefore, the
equation of the line is y = x − 15.
Pros: We only have to use the slope-intercept form, which we should be very familiar with by now.
We do not have to use distribution, and the solving is typically easier.
Cons: Once we plug everything into the equation and solve for the y-intercept, we have to remember
to rewrite the equation with the x and y still intact.
Example 1c (Mental Method)
Determine a linear equation with a slope of that contains the point (-6,4).
For this method we will use the slope-intercept form of a line. We know that the equation will have
the form of y = x + ___, where the blank represents some value. We ask ourselves “What is times
-6?” and the answer is -10. -10 obviously does not equal 4, so how do we get from -10 to 4? We
have to add 14. Which means we can now fill in the blank and get y = x + 14.
Pros: No rewriting, no simplifying, and no solving are required.
Cons: We have to be able to mentally multiply and visualize a “number line”.
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An extra step is necessary if we are given two points without a slope. If given two points, we will have
to calculate the slope and then use one of the two points to develop a regression equation.
Example 2 (Mental Method)
Calculate a linear regression equation that intersects the points (3,6) and (9,-2).
Since the y-values decrease from left to right, the slope is negative. The rise is 8 and the run is 6.
Therefore the slope is − . Then all we have to do is choose a point. Either point will result in the
same linear equation. − times 3 is −4. To get from -4 to 6, we will have to add 10. − times 9 is
-12. To get from -12 to -2, we will also have to add 10. Therefore both points lie on the line whose
equation is y = − x + 10.
Applications
Very often, application problems (better (actually worse) known as WORD PROBLEMS) of a linear
nature involve the creation of a regression equation to determine interpolations and extrapolations,
and these questions are the meat and potatoes of almost all standardized tests.
Example 3
A national cell phone service is charging $50 per month plus 13 cents for every text message. If on
average Harry sends 3500 texts per month, how much should Harry’s cell phone bill be for the
month?
In this case if he sends no text messages, his bill would be $50. This is essentially the y-intercept.
Additionally, his rate is 13 cents per text, which is the slope. The equation for this problem is
y = .13x + 50, where x is the number of minutes and y is the total in dollars. If we let x = 3500, then
y = .13(3500) + 50 = $505.00. At which point Harry gets kicked out of his house and ends up eating
twinkies out of someone’s garbage.
Example 4
Harry’s parents forgive him, but they switch him to a new plan. The first month on the new plan,
Harry only sends 1000 texts, and his bill is $90. The second month he sends 1500 texts and his bill is
$100. Based on his two bills, what is the new plan charging per month, how much is it charging per
text, and how much will his bill be if he returns to texting 3500 texts per month?
Since his bill increases by $10 for an increase of 500 texts, the rate of change is
=
= $0.02 per
text. If we multiply
(1000) = = $20. If his bill is $90, and $20 of the bill was a result of texting,
then he is being charged $70 per month. We get the equation y = x + 70. If he sends 3500 texts,
then we get y = (3500) + 70 = $140. Harry’s parents would be slightly disgruntled, but they will
probably allow him to eat twinkies out of the pantry.
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I.B.4 Independent Practice
Determine linear equations for the following slopes and points.
1. Slope = 5; (6,-3)
6. Slope = ; (6,-2)
2. Slope = -2; (-5,2)
7. Slope = − ; (10,7)
3. Slope = 8; , −3
8. Slope = ; (6,-5)
4. Slope = 6; , 7
9. Slope = ; , 2
5. Slope = 8; , 4
10. Slope = − ; , −4
Determine linear equations for the following points.
11. (3,5),(6,9)
12. (5,2),(8,-4)
13. (-3,8),(2,-2)
14. (-5,7),(-9,-1)
15. (6,3),(4,11)
16. (3,9),(9,5)
17. (6,8),(4,5)
18. (5,3),(-10,9)
19. (2,7),(5,5)
20. (3,8),(7,14)
21. Gerard can eat 5 pickles every hour. If he has already eaten 12 pickles, how many pickles will
he have eaten after 6 hours?
22. Gerard has improved his pickle eating prowess. If after 8 hours he has eaten 72 pickles, and
after 12 hours, he has eaten 102 pickles, how many pickles can he eat every hour? How
many pickles had he already eaten when he started counting? How many pickles will he have
eaten after 24 hours?
23. Gerard is in the hospital. The doctors have given him medicine, which purges the pickles from
his body at a constant rate. If after 2 hours he has the equivalent of 184 pickles still in his
system, and after 11 hours he still has 112 pickles in his system, how long will it take for him to
purge all of the pickles from his system?
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Section
I.B.5
Parallel and
Perpendicular
Parallel lines are defined as coplanar lines that do not intersect. The coplanar is an extremely
important qualifier. Without it, non-intersecting lines could be skew, which are defined as noncoplanar
lines. Perpendicular lines are defined as lines that intersect at a right (90°) angle. (If two
lines intersect, they are automatically coplanar)
Myth: If two coplanar lines are not parallel, then they are perpendicular.
Fact: Because the terms parallel and perpendicular are often used in tandem with one another, many
students consider them opposites. Perpendicular is a very specific type of intersection, and, thus,
only a small portion of what many consider “The opposite of parallel”.
Parallel and Perpendicular in Geometry
Parallel and perpendicular lines are everywhere in geometry. To best understand them we will need
to define a few terms. Firstly, a right angle is is equivalent to the corner of a square. We say that it
has a measure of 90°. This is an arbitrary value derived from the Mayan Calendar (III.A.2), and the
concept of degree is not mathematically viable for future courses. In Unit IV, we will investigate a
new method for calculating angle measures, which can be used in courses like calculus. An acute
angle is any angle smaller than a right angle, and an obtuse angle is any angle larger than a right
angle. A line is sometimes coined a straight angle with an angle measure of 180°. Two angles that
share a side and a vertex are considered adjacent, and two adjacent angles that form a line with their
non-shared sides are called a linear pair, and since their measures sum to 180°, we call them
supplementary. Finally, a transversal is a line that intersects two or more parallel lines.
When a transversal intersects parallel lines, all acute angles have equal measure as do all obtuse
angles. Each acute angle is supplementary to each obtuse angle. Parallel lines also pop up in
parallelograms and trapezoids. Since a parallelogram is essentially made up of two pairs of parallel
sides that act like transversals to each other, the acute angles are supplementary to the obtuse
angles.
Even though parallel lines take the spotlight for much of geometry, perpendicular lines are much more
powerful. Perpendicular segments and lines intersect in the corners of squares and rectangles, in the
diagonals of squares and rhombi, in the radii and tangent lines of circles, and most prominently in the
legs of the aptly named right triangles. Right triangles are the basis for both Units III and IV, so we
will not spend too much time on them here, however it should be interesting to note, that this is the
only section in this text that really discusses parallelism, while perpendicularity takes up two whole
units.
Parallel Linear Equations
The most important fact about parallel lines that we discuss in Algebra is the fact that two parallel
lines have the same exact slope. Usually, this is the only significance given to parallel lines before
we start encountering problems, however, if we think of the applications from the previous section,
identical slopes mean equal rates of change. So if two phone company A charges the same rate for
texting as company B, but company A’s monthy charge is $20 more, then a person using company A
will always have a larger bill than company B.
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Example 1
Calculate a linear equation for the line parallel to y = x + 2 that contains (10,3).
Since the two lines are parallel, the new line must also have a slope of . (10) = 6, so we have to
subtract 3 to get to a y-value of 3. Therefore the linear equation is y = x − 3.
Perpendicular Linear Equations
Two lines that are perpendicular have slopes that are negative reciprocals of each other.
Geometrically, this is extremely interesting and will aide us in transformations later on. Applicationwise,
this literally means nothing besides the fact that one line will have a positive rate of change and
the other will have a negative rate of change.
Myth: A number and its reciprocal have opposite signs.
Fact: A reciprocal is merely a flipped fraction. There is never a sign change. Unfortunately, due to
the focus on perpendicularity, the term “negative reciprocal” becomes synonomous with “reciprocal”
for most students.
Example 2
Calculate a linear equation for the line perpendicular to y = x + 2 that contains (12,4).
As the two lines are perpendicular, the second line will have a slope of − . − (12) = −20 to which
we must add 24 to get to y = 4. Thus the equation is y = − x + 24.
Standard Form Equations
If two lines are parallel, their standard form equations will be of the form Ax + By = C and
Ax + By = D where A, B, C, and D are all integers. It is apparent that the two equations have
identical left sides, which is a result of the two lines have identical slopes.
If two lines are perpendicular, their standard form equations will be of the form Ax + By = C and
Bx − Ay = D where C and D may or may not be equal. It is apparent that the coefficients have
switched and the sign between them have changed. As slopes are y-values over x-values, the
reciprocal switches the x and y-values. We also have to negate either the x-value or the y-value, but
since generally the x-coefficient is positive, the y-value gets negated.
Example 3
Calculate a linear equation for the line perpendicular to 3x + 4y = 3 and contains (-3,6).
The left side will be of the form 4x − 3y, and if we plug in the given point 4(−3) − 3(6) = −30.
Therefore the equation will be 4x − 3y = −30.
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I.B.5 Independent Practice
For the following equations and points, calculate linear equations for both the parallel line and
perpendicular line that contain that point.
1. y = 3x + 2; (6,4)
2. y = 5x − 2; (10,3)
3. y = x + 5; (6,3)
4. y = −3x + 8; (−9,2)
5. y = −4x − 2; (−12, −3)
6. y = − x + 7; (15, −4)
7. y = x + 9; (6,5)
8. y = x + 10; (−15,32)
9. y = − x − 5; (5,3)
10. y = x + 3; (5,7)
11. 3x + 5y = 18; (4,7)
12. 5x + 6y = 21; (−3, 2)
13. 4x + 7y = 13; (5, −2)
14. 6x − 2y = 5; (3,4)
15. 3x − 5y = 12; (−2, −7)
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Chapter
I.C
59
Cogswell – PCM

Chapter
I.C Chapter
Overview
‣ I.C.1 – Distribution and Polynomial Multiplication
This section investigates the distribution of monomials across polynomials. Additionally, it will
cover the multiplication of polynomials with the main focus being on binomial multiplication
resulting in a quadratic equation.
‣ I.C.2 – Factoring
This section reverses the distribution covered in the previous section through the concept of
factoring. Methods used will be the box method, factoring by grouping, and mental factoring.
‣ I.C.3 – Solving a Quadratic Equation
This section will utilize factoring to solve quadratic equations. It will also utilize the quadratic
formula to solve non-factorable equations. Calculator operations involving storing values and
utilizing lists will be introduced to ease the process of the quadratic formula.
‣ I.C.4 – x-intercepts and y-intercepts
This section directly ties to I.B.4, and shows how to find both x and y-intercepts for a parabola.
‣ I.C.5 – Graphing Vertex Form
This section allows the conversion from general form to vertex form as well as how to graph a
parabola efficiently.
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Cogswell – PCM

Section
I.C.1
Distribution and
Polynomial Multiplication
Back in Section I.A.4, we discussed the concept of distribution. When multiplying a factor by a sum,
we can multiply the factor by each individual term within the sum. In other words:
a(b + c) = ab + ac
We are not limited to distributing through binomials. In fact we can distribute through any
polynomial. The concept is the same either way. Additionally, division of a polynomial by a single
term can be rewritten as distribution.
= (b + c) = +
However, we cannot divide a term by a polynomial. Doing so rips a hole in the space-time
continuum… or results in a terribly wrong answer… one of the two definitely happens…
Example 1
Simplify 5x y (2x + 3y).
≠ +
5x y (2x + 3y) = 5x y (2x) + 5x y (3y) = 10x y + 15x y
Example 2
Simplify 3x (5x − 2x + 7x − 2).
3x (5x − 2x + 7x − 2) = 3x (5x ) + 3x (−2x ) + 3x (7x) + 3x (−2) = 15x − 6x + 21x − 6x
Example 3
Simplify
.
=
(6x + 5x − 2) =
+
−
= 3x + −
As we can see, distribution is an extremely useful process, as long as it is used correctly. Its true
power can be seen when multiplying polynomials together. Back in Algebra 1, many students learned
a concept referred to commonly as FOIL. FOIL is an acronym that stands for First, Outside, Inside,
Last, the order in which we multiply two binomials together. This is all well and good until we are
presented with a trinomial or polynomials with more terms. Thankfully, we can use a process we will
refer to as Extended Distribution.
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Extended Distribution
Extended Distribution is best illustrated through example. Essentially, we will distribute the second
polynomial across the first polynomial and then distribute each term from the first polynomial across
the second polynomials. Again, it is best illustrated through example. We were warned.
Example 4
Simplify (3x + 5)(2x − 7).
(3x + 5)(2x − 7) = 3x(2x − 7) + 5(2x − 7) = 3x(2x) + 3x(−7) + 5(2x) + 5(−7)
= 6x − 21x + 10x − 35 = 6x − 11x − 35
To be fair, we could have used FOIL to achieve the same answer, but through extended distribution
we can perform the process with limited written work. Plus we can simplify the following.
Example 5
Simplify (5x − 2)(3x + 4x − 9).
(5x − 2)(3x + 4x − 9) = 5x(3x + 4x − 9) − 2(3x + 4x − 9) = 15x + 20x − 45x − 6x − 8x + 18
= 15x + 14x − 53x + 18
Furthermore, FOIL never allowed for multiplying three binomials.
Example 6
Simplify (3x + 2)(4x − 5)(7x + 2).
(3x + 2)(4x − 5)(7x + 2) = (3x + 2)(28x + 8x − 35x − 10) = (3x + 2)(28x − 27x − 10)
= 84x − 81x − 30x + 56x − 54x − 20 = 83x − 25x − 84x − 20
Important Note: A negative sign in front of a binomial is essentially a -1. Therefore
Example 7
Simplify – (3x − 2)(2x + 5).
– (ax + b) = −ax − b and – (ax − b) = b − ax.
−(3x − 2)(2x + 5) = (2 − 3x)(2x + 5) = 4x + 10 − 6x − 15x = −6x − 11x + 10
The negative sign only distributes across one binomial, never both.
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I.C.1 Independent Practice
Simplify the following.
1. 4(3x + 6) =
11. (2x − 5)(2x − 4) =
2. 5x(3x − 2) =
12. – (5x − 2)(6x + 5) =
3. 4x (3x − 1) =
13. −3(2x − 5)(3x − 7) =
4. 5x y (3xy − 8x ) =
14. (6x + 2)(9x − 5x + 1) =
5. 3x(4x + 5x − 2x) =
15. (x − 5)(3x − 4x − 8) =
6. 5x y (5x − 8xy + 4y − 3) =
16. – (2x + 3)(4x + 5x − 2) =
7.
=
17. (3x + 5x − 1)(7x + 4x − 8) =
8.
=
18. −2(4x − 2x + 7)(2x + 5x − 6) =
9. (3x + 5)(6x − 2) =
19. (2x + 5)(2x − 4)(6x + 2) =
10. (4x + 2)(8x + 1) =
20. (8x + 2)(2x − 6)(3x + 5x − 1) =
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Cogswell – PCM

Section
I.C.2 Factoring
Ah factoring… the third and most fearsome of all mathematical f-words. This is the one concept that
truly causes anxiety in most students. Whereas fractions are disliked and functions are
misunderstood, factoring is sincerely hated with a passion equal to the heat of a thousand suns by
almost everyone. We may be over-stating it but not by much.
Why is there anxiety? Teachers have not always been teachers. Once they were students just like
us. When teachers were taught factoring, many of them were taught with the guess-and-check
method. For anyone who loves mathematics or may deep down have the potential to love
mathematics, guess-and-check feels wrong. There is something innately frustrating about doing a lot
of work only to get the wrong result and having to start over, not because we made a mistake but
because we guessed wrong. Plus, wasting paper is bad for the environment, right? These future
educators disliked factoring when they learned it, and so feel very uncomfortable teaching it to their
students. The trend is likely to repeat itself unless we do something about it.
Methods for factoring have been developed to eliminate the need for guessing and checking, but
many of them are simply guess-and-check in disguise. Others have a little guess-and-check in them.
Some require a grid or other pictorial device, and still others require no writing at all. There are even
a few that require “fake math” to perform. How do we choose? We need a sure-fire way of factoring
that works in all cases, is easy to do, and eliminates all guess work. It would be nice if the method
were mathematically sound as well. Regardless of what method we feel most comfortable with, all
factoring boils down to one simple concept, distribution, or, in the case of factoring, “reverse”
distribution.
We associate the word distribution with multiplication, because we multiply the factor by each term in
the polynomial. When we are told to factor something, we are being asked to reverse the distribution
process. In other words, factoring implies division. Amazingly enough, two of the mathematical f-
words mean division.
Example 1
Factor 3x + 6x.
The two terms in the binomial have one x in common. Also the 3 and the 6 can both be divided by 3.
So, we can divide (or factor) out a 3x. 3x + 6x = 3x
= 3x(x + 2).
Once we make peace with the fact that factoring is division, we can eliminate some of the steps in
factoring.
Example 2
Factor 5x + 10x − 20x .
5x + 10x − 20x = 5x (x + 2x − 4)
64

Reverse Extended Distribution
Reverse extended distribution is also known as factoring by grouping, but since it is the exact
opposite of the process we saw in the previous section, it makes more sense if we think of it as
extended distribution in reverse.
If we multiply (3x + 2)(x − 5), we get 3x − 13x − 10. The −13x comes from the combining of −15x
and 2x. If we know that −13x needs to be divided up into −15x and 2x, then we can easily go
backwards. 3x − 13x − 10 = 3x − 15x + 2x − 10 = 3x(x − 5) + 2(x − 5) = (3x + 2)(x − 5). The
trick is figuring out how to effectively divide −15x into usable terms. If we multiply the first term, 3x ,
by the last term, −10, we get −30x . We need two factors of −30x that combine to equal −13x.
The only two terms that satisfy this situation are −15x and 2x, the very terms we need to perform
reverse extended distribution.
Example 3
Factor 5x + 12x + 4.
Multiplying 5x by 4, we get 20x . What two factors multiply to get 20x and combine to get 12x?
The terms are 10x and 2x. Knowing this we can reverse distribute.
5x + 12x + 4 = (5x + 10x) + (2x + 4) = 5x(x + 2) + 2(x + 2) = (5x + 2)(x + 2)
Example 4
Factor 4x + 5x − 6.
(4x )(−6) = −24x = (8x)(−3x)
4x + 5x − 6 = (4x + 8x) + (−3x − 6) = 4x(x + 2) − 3(x + 2) = (4x − 3)(x + 2)
Example 5
Factor 4x − 9.
(4x )(−9) = −36x = (6x)(−6x)
4x − 9 = (4x + 6x) + (−6x − 9) = 2x(2x + 3) − 3(2x + 3) = (2x − 3)(2x + 3)
Example 6
Factor 4x − 40x + 100.
4x − 40x + 100 = 4(x − 10x + 25) = 4(x − 5x) + (−5x + 25) = 4x(x − 5) − 5(x − 5)
= 4(x − 5)(x − 5) = 4(x − 5)
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Cogswell – PCM

I.C.2 Independent Practice
Factor the following fully.
1. 25x + 40x
11. 100 − 36x
2. 49x − 14x
12. x − 49
3. 3x y + 18x y
13. 2x + 54x + 100x
4. 8x y z − 24x y z
14. 6x + 15x − 9x
5. x + 15x + 36
15. 4x − 24x + 11
6. x − 13x − 48
16. 16x − 12x − 10
7. 3x − 10x + 8
17. 24x + 18x − 27x
8. 4x + 16x − 9
18. 15x − 14x + 3
9. 16x + 24x + 9
19. 9x + 25x − 6
10. 9x − 30x + 25
20. 10x + 27x + 14
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Section
I.C.3
Solving a Quadratic
Equation
There are multiple ways to solve a quadratic equation, but there are really only two ways that are
necessary: factoring and the quadratic formula. Yes, the quadratic formula works for every problem.
However, there are times when we will be expected to solve a quadratic equation by hand, and
factoring will be the easier alternative. Quadratic equations can have no solutions, one solution, or
two solutions. We are not going to worry about imaginary numbers.
Solving by Factoring
The two keys to solving by factoring are knowing how to factor and understanding that if a product is
equal to zero, then one of the factors has to be zero. The first step is to have one side equal zero,
then we factor, set each factor equal to zero, and solve each new linear equation separately.
Example 1
Solve 4x − 21x + 5 = 0 for all values of x.
4x − 21x + 5 = 0
(4x − 1)(x − 5) = 0
4x − 1 = 0 or x − 5 = 0
x = or x = 5
Example 2
Solve 3(x + 2)(x − 1) = x + 2 for all values of x.
3(x + 2)(x − 1) = x + 2
3(x + x − 2) = x + 2
3x + 3x − 6 = x + 2
3x + 2x − 8 = 0
(3x − 4)(x + 2) = 0
3x − 4 = 0 or x + 2 = 0
x = or x = −2
Example 3a
Solve 4x + 7x − 5 = 0 for all values of x.
4x + 7x − 5 does not factor. We cannot solve this using factoring.
67

Solving with the Quadratic Formula
For problems that cannot be factored, the quadratic formula is absolutely essential. This formula is
one of the few formulas we must memorize, because going about these problems the long way takes
much longer.
Example 3b
Solve 4x + 7x − 5 = 0 for all values of x.
If ax + bx + c = 0, then x = ±√
In this quadratic a = 4, b = 7, and c = −5. Thus, x = ± ()()
()
The Quadratic Formula and the Calculator
= ±√
= .545 and − 2.295
To utilize the quadratic formula in the calculator, we will make use of two useful functions: storing and
lists. Firstly, we will store the a value into A, the b value into B, and the c value into C, using the STO
button. Then we will type (−B + {1, −1} √(B^2 − 4AC))/(2A). We must put the entire numerator and
denominator inside their own parentheses because the calculator works with order of operations and
we need it to calculate them separately before dividing. The {1,-1} stands for ± in the calculator, and
creates a list with all answers. If we receive an error ERR:NONREAL ANS, there are no solutions to
the quadratic equation.
Example 4
Solve 3(x + 4) = 6 for all values of x.
3(x + 4) = 6
3(x + 8x + 16) = 6
3x + 24x + 48 − 6 = 0
3x + 24x + 42 = 0
‘3’ ‘STO’ ‘A’ ‘ENTER’
‘24’ ‘STO’ ‘B’ ‘ENTER’
‘42’ ‘STO’ ‘C’ ‘ENTER’
We then type: (−B + {1, −1} √(B^2 − 4AC))/(2A) and get -2.586 and -5.414 as our two answers.
Example 5
Solve 4x + 5x + 8 = 0
Solving this in the same way as Example 4, we get an error, which indicates there are no real
solutions to this equation.
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I.C.3 Independent Practice
Solve the following by factoring.
1. 3x − 28x + 9 = 0
4. 4(x − 5) = 3x − 11x
2. 4x + 10x − 6 = 0
5. 7x(x + 3) = x + 20x + 2
3. 2x − 15x − 8 = 0
6. 5x(x + 3) + 9 = −3x
Solve the following with the Quadratic Formula. x = ±√
7. 7x + 5x − 2 = 0
12. 7(9x + 4x) = 4x + 5
8. 3x − 4x + 5 = 0
13. 4x(4x − 5) = 32
9. 7x − 4x + 5 = 0
14. 5(x + 6) = 7x − 2
10. 4x − 6x + 2 = 0
15. 4(x − 3)(x + 5) = 2x − 5
11. 4x + 5x − 8 = 0
16. −3x(x + 5) = (2x − 2)
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Section
I.C.4
x-intercepts and
y-intercepts
In I.B.3, we discussed the concept of x and y-intercepts. Conceptually nothings has changed, but
because we are now dealing with quadratic equations, solving for x-intercepts and y-intercepts
becomes slightly more involved. To find the y-intercept (quadratics only have one), we substitute 0
for x and solve for y, just like we did before. To find the x-intercept(s), or root(s), we substitute 0 for y
and solve for x.
Quadratic functions, like linear functions, can take on several forms. The general form is how we
refer to y = ax + bx + c. The other common form, vertex form, looks like y = a(x − h) + k, where
(h, k) is the vertex of the parabola. The vertex is the highest point if the parabola is concave down
and the lowest point if the parabola is concave up. We will discuss vertex form more at length in the
next section, but we will practice with it in this section. Either way, quadratics have x in them.
Example 1
Find the x-intercept(s) and y-intercept of y = 4x − 11x + 6.
Finding the y-intercept is easier, so we will find it first. Substituting 0 for x, we get
y = 4(0) − 11(0) + 6 = 6. Therefore the y-intercept is (0,6). Substituting 0 for y, we get
0 = 4x − 11x + 6 = (4x − 3)(x − 2). Therefore the x-intercepts are , 0 and (2,0).
Example 2
Find the x-intercept(s) and y-intercept of y = (x − 3) + 5.
Again, substituting 0 for x, we get y = (0 − 3) + 5 = (−3) + 5 = 9 + 5 = 14. The y-intercept, is
thus, (0,14). Substituting 0 for y, we get 0 = (x − 3) + 5. We can solve this like a normal fraction
without factoring. 0 = (x − 3) + 5 → −5 = (x − 3) → √−5 = x − 3. Since we cannot find the square
root of -5, there are no x-intercepts.
Example 3
Find the x-intercept(s) and y-intercept of y = 9x + 30x + 25.
y = 9(0) + 30(0) + 25 = 25. Therefore (0,25) is the y-intercept.
0 = 9x + 30x + 25 = (3x + 5)(3x + 5) = (3x + 5) . Therefore − , 0 is the only x-intercept.
Alternately, we could also use the quadratic formula: x = ± ()()
= ±√
= −
= − .
()
Example 4
Find the x-intercept(s) and y-intercept of y = −2(x − 5) + 8.
−2(0 − 5) + 8 = −42; y-intercept: (0, −42).
0 = −2(x − 5) + 8 → 4 = (x − 5) → ±2 = x − 5 → x = 5 ± 2 = {7,3}; x-intercepts: (7,0), (3,0).
70

I.C.4 – Independent Practice
Find the x-intercept(s) and y-intercept for each quadratic below.
1. y = 3x − 22x + 7
5. y = 14x + 24x + 10
2. y = 4x + 8x − 5
6. y = −4x − 13x + 12
3. y = 5x + 36x + 7
7. y = 6x − 22x + 20
4. y = 2x − 3x − 9
8. y = −8x + 26x + 7
9. y = (x − 5) + 7
13. y = 2(x + 6) − 50
10. y = (x + 2) − 16
14. y = −3(x − 4) + 75
11. y = −(x − 3) + 36
15. y = 5(x + 2) − 15
12. y = −(x + 5) − 14
16. y = −8(x − 7) + 16
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Section
I.C.5 Vertex Form
Why in the world do we need to know more than one form of a quadratic equation? The truth is that
the vertex form of a quadratic function, y = a(x − h) + k, is much more useful than the general form,
y = ax + bx + c. However, the general form is more prominent and more aesthetically pleasing to
the eye. One of the major strengths of vertex form should be evident after completion of I.C.4: mainly
the ease of determining x-intercepts algebraically. Where the general form requires factoring or the
quadratic formula, the vertex form requires only that we can solve an equation. This is due in part to
the fact, that it has pretty much already been factored for us. How nifty is that? The other strength of
vertex form is the quick identification of the vertex, namely (h, k), as well as determination of its
concavity and vertical dilation factor.
Sketching a Graph Based on Vertex Form
The vertex, (h, k), is the point at the top or the bottom of the parabola. The leading coefficient, a, has
a much more important duty. If positive, we say the parabola is “concave up”, and if negative, we say
the parabola is “concave down”. If −1 < a < 1, the parabola is vertically condensed, which (as a
byproduct) appears to widen the parabola. If a > 1 or a < 1, the parabola is vertically stretched,
which appears to thin the parabola. Again, we are simply going to systematically choose several x-
values, determine their corresponding y-values, and plot the resulting coordinates.
Example 1
Sketch a graph of f(x) = 2(x − 3) + 5.
We know (3,5) is the vertex, so now let us choose two values to the left and
two to the right, in this case {1,2,4,5}. f(1) = 13, f(2) = 7, f(4) = 7,
f(5) = 13. If we plot these points and connect them we will get the graph to
the right.
Example 2
Sketch a graph of f(x) = − (x + 2) + 3.
We know (−2,3) is the vertex. So we should choose x-values on both
sides of −2. Additionally, since our leading coefficient is a fraction with 2 in
the denominator, it would be wise to select x-values by counting by twos:
{−6, −4,0,2}. f(−6) = −5, f(−4) = 1, f(0) = 1, & f(2) = −5. We can
then plot these points and connect them.
y
x
This is all well and good, but what happens if the quadratic equation facing us is a general form
equation? How do we find the vertex? How do we graph an equation we know nothing about?
72

Converting to Vertex Form
So the main question we should now ask ourselves, is how do we convert a general form quadratic
equation into vertex form? Well, there are two schools of thought on this. The first believe in a
process called “Completing the Square”, which is useful when finding the equations of all conics, like
circles and ellipses. The second believe that conics, not being terribly useful in life, are not a good
enough reason to force students to undergo the torture they usually experience completing squares.
We will experience this torture when we get to circles, but now is just not a good time. When
converting from ax + bx + c to a(x − h) + k, the a stays the same, and h = − , a modified form of
the quadratic formula.
Example 3
Convert f(x) = 3x + 12x − 4 into vertex form.
h = −
() = −2. k = f(−2) = 3(−2) + 12(−2) − 4 = −16. Therefore the vertex form is
f(x) = 3(x + 2) − 16.
Example 4
y
Sketch a graph of f(x) = x + 4x + 6.
h = −
= −4. k = f(−4) = −2. The vertex form is f(x) =
(x + 4) − 2 with a vertex at (−4, −2). Picking four other points
f(−8) = 6, f(−6) = 0, f(−2) = 0, & f(0) = 6. We can now sketch the
graph.
x
As we can see, quadratic equations are more complicated to graph than linear equations, but the
basic idea of plotting points that satisfy the equation to get a general sketch holds true no matter the
difficulty of the equation. As we get into new types of more complex equations, we have to keep this
in mind.
This space intentionally left blank for soda stains. NO TEA!!!
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I.C.5 Independent Practice
Sketch a graph of each of the following.
1. y = 3x − 5
y
1.
2.
x
y
x
2. y = − x + 3
y
3.
4.
y
3. y = x − 4x + 3
x
x
4. y = 2x − 12x + 14
5. y = −3x + 12x − 4
y
5. 6.
y
x
x
6. y = – x − 10x − 19
7. y = x + 2x − 3
y
7.
8.
y
x
x
8. y = x − 8x + 2
9. y = − x − 4x − 5
9.
y
10.
y
x
x
10. y = − x + 40x − 156
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Unit
II
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Cogswell – PCM

Unit
II Unit Overview
The Purpose of the Unit
The purpose of this unit is to build upon the algebraic foundation laid out by Unit I. Some of it may be
a review of concepts from Algebra 2. Most of it will be new concepts to help further algebraic
proficiency. We will investigate new types of functions beyond linear and quadratic, as well as
develop an understanding of function manipulation, transformation, and inversion. We also will focus
on function division, using synthetic division, to better understand rational functions. Furthermore, we
will solve systems of linear, quadratic, power, and exponential functions, algebraically and
graphically.
This unit will develop our understanding of matrices and how to use them in solving systems of multivariable
equations and transforming graphs by rotation, transformation, and dilation. Finally, this unit
will end with the conceptualization and calculation of arithmetic, geometric, and power sequences and
series, a field of mathematics which will be completely new.
The Advancements of Mathematical Technology Throughout History
If we ever venture into a dollar store and walk down the school supply aisle, invariably we will see
calculators with all ten digits, +, −, ×, ÷, √ , %, & =. These come in a variety of super exciting colors
and are commonly attached to rulers, binders, or key chains. With the increased usage of scientific
and graphing calculators, many times we overlook these four-function gems. Most cell phones come
fully equipped with one of these as well. What is amazing is the fact that computers were initially
developed in order to automate these basic functions in a time-efficient way. These initial computers
filled entire rooms and, in some cases, entire floors with metallic boxes, switches, lights, and tape
reels. Now 5-year-olds can get a four function calculator as a party favor, not that it would be
appreciated. After all there is candy to be had.
The development of an affordable 4-function hand-held calculator did not come to fruition until the
early 1980s. Expensive scientific calculators would appear in the mid-eighties with advanced
functions like logs, sines, and cosines. Two-hundred-dollar graphing calculators would debut in the
nineties with limited capabilities. So what did scientists and mathematicians use before calculators
became commonplace?
Advanced mathematicians calculated by hand and measurement a multitude of mathematical values,
and these were listed in tomes filled with tables. The slide-rule was also developed as a mechanical
calculator, some of the most advanced ones able to do calculations in less time than it takes to input
the information into today’s calculator. If we ask our parents and grandparents about their
mathematical education, we will experience groans from memories of slide rules, compasses, and
hand cramps from hand-calculation. Thanks to advances in technology, we do not have to
experience these issues, but it should make us appreciate what has come before and what we have
now.
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Chapter
II.A
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Chapter
II.A Chapter
Overview
‣ II.A.1 – Exponential Functions
This section investigates the effect of a variable in the exponent. We will distinguish between
exponential and power functions and look at their graphs. We will also use the regression
operations to determine equations based on given points.
‣ II.A.2 – Logarithmic Functions
This section investigates the true meaning behind logarithmic functions as inverses of exponential
functions. We will investigate the natural number, e, and the natural logarithmic function. We will
look at the logarithmic graph and its relationship to the exponential graph.
‣ II.A.3 – Function Division
This section will introduce polynomial long division as well as synthetic division in preparation for
sections II.A.4 and II.A.5.
‣ II.A.4 – Polynomial Functions
This section will continue the pattern set by linear and quadratic functions. We will expand the
concept of domain and range, as well as discuss end behaviors and the functions’ natures at the
x-intercepts.
‣ II.A.5 – Rational Functions
This section will introduce rational functions, vertical asymptotes, horizontal asymptotes, and
holes. We will graph these discontinuous functions as well as use elementary limits to investigate
the functions’ natures near discontinuities.
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Section
II.A.1 Exponential
Functions
Back in I.A.3, we discussed the rules for multiplying and dividing exponents. The same rules still
apply, however now we are going to apply our general exponential knowledge to exponential
functions. Firstly, exponential functions are of the form f(x) = ar , where (a, 0) is the y-intercept and
r is considered the common ratio, which is technically similar to the slope in a linear equation. The
common ratio is the value we multiply each consecutive term by.
Finding an Exponential Function
If we are given two points, (2,8) and (3,16), we see that an increase by 1 in the x-values results in the
doubling of the y-values. Therefore, in this case, the common ratio is 2. If however, we have two
points, (2,5) and (4,45), we see that an increase by 2 in the x-values results in multiplying the y-value
by 9. What has actually occurred is multiplying by 3 twice, which actually makes the common ratio 3
in this case.
We could use a formula and simplify it to get the exponential function:
Example 1
f(x) = y (r)
Determine an exponential function that contains the points (2,5) and (4,80).
An increase of 2 results in a multiplication of 16, which means the common ratio is √16 = 4.
Therefore the function is f(x) = 5(4) = 5
= (4) . We could also have used the other point
f(x) = 80(4) = 80
=
(4) = (4) .
Example 2
Determine an exponential function that contains the points (3,81) and (6,3).
An increase in 3 results in a multiplication of . Which means the common ratio in this case is
= . Therefore the exponential function is f(x) = 81 = 81 (3 ) = 2187 .
Graphing an Exponential Function
We also need to be able to graph an exponential function. Like graphing a line or a parabola, we
need to choose x-values wisely, determine their corresponding y-values, and plot the points to give us
a general skeleton of the graph itself. To graph an exponential, we really only need three points: the
y-intercept and two others. If r > 1, we will find the two points to the right, and if r < 1, we will find
the two points to the left. Additionally, all exponential functions have a horizontal asymptote on the
x-axis.
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Example 3
Sketch the graph of the function y = 3(2) .
We know that the y-intercept is (0,3), and since the common ratio, 2, is greater
than 1, we will use the x-values 1 and 2. Finding the corresponding y-values,
we find the points (1,6) and (2,12). Graphing these three points results in the
graph at right.
y
x
We should notice as the x-values get lower, the graph gets closer to the x-axis, but they never
intersect. This is an example of exponential growth.
Example 4
Sketch the graph of the function y = 2 .
y
The y-intercept is (0,2). Choosing the two values to the left gives us (-1,6) and
(-2,18). The graph is shown at right.
In this case as the x-values increase, the y-values get closer and closer to the
x-axis. This is an example of exponential decay.
Exponential Growth and Decay
x
Exponential growth and decay constitute the most useful application of exponential functions. We
see these used in finance, biology, chemistry, and atomic physics.
Example 5
A given population of a Texas town is increasing by 12% yearly. If the population started out at 1500
people, how many people will there be in 10 years?
In this case, we have a percent increase, which means our common ratio is 1.12. We also have been
given the y-intercept of (0,1500). Therefore the equation is y = 1500(1.12) . If we, thus, substitute
10 for x, we will get 4658.77, or roughly 4659 people.
Example 6
A certain isotope has a half-life of 15 years. If we start out with 100 grams, how much will be left after
50 years?
Because half the material is left after 15 years, the common ratio is . However, we will also have to
divide the number of years that pass by 15 to get the number of half-lives. Therefore, the equation
will be y = 100 . Plugging in 50, we get 9.921 grams.
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Independent Practice II.A.1
Determine an exponential equation that contains the following points.
1. (3,5) & (4,25)
2. (0,35) & (1,5)
3. (0,6) & (2,54)
4. (3,18) & (5,2)
5. (0,6) & (3,48)
6. (2,128) & (5,2)
Determine three points that would allow for the graphing of the following equations.
7. y = 5(3)
8. y = 24 9. y = 7(5)
10. y = 20
Solve the Following Application Problems
11. Monica decides to invest her $3000 in a fund that compounds annually at an interest rate of
7%. How much will she have after 18 years?
12. A bacteria culture started with 10 cells. If after two hours, the number of cells have tripled, how
many cells will there be in 24 hours?
13. Freddy wins a staring contest with his sister. He says she can pay him $30 now, or she can
pay him a penny now, doubling the amount each day for the next two weeks. Should she take
the deal?
14. A virus has a half-life of 3 hours. If Dawn has 1,000,000,000 virus cells in her body at
midnight, how many will she have in her body in 24 hours?
15. Phillip breeds rabbits. He starts with two rabbits, and in three years, he has 32 rabbits. How
many rabbits will he have in eight years?
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Section
II.A.2 Logarithmic
Functions
Firstly, those students who have never worked with logs are currently at an advantage because logs
are taught incorrectly in almost every school in the United States. Okay, so that fact can’t be
confirmed without a math census, but we will assume that the statement is true. Those students who
have been taught logs in the past are now going to actually learn them.
Let’s begin by examining how they are taught in most schools. For this section, feel free to read it
aloud in a screechy, high-pitched old woman voice or a monotone, extremely low old man voice.
The OLD Method
“Now that we all understand exponential functions, what happens if the variable is the exponent? We
use something called logarithmic functions to solve this. Firstly, we have to learn how to rearrange an
exponential function so that we can solve it. Here is a diagram of how the problem should be
rearranged.” Without fail, this is the diagram displayed.
“Now, here is a worksheet for you to rearrange thirty exponential equations.”
Two to three days later after much rearranging, students are given the properties (which are hard to
remember) and how to solve an exponential equation with a calculator (after using yet another
property). Two weeks after these lessons, a typical student has completely forgotten how to solve
these types of problems.
The NEW Method
b = a
log a = x
Please don’t use a funny voice to read this section, unless, of course, it’s a really funny voice. Before
we begin let’s investigate the concept of inverse function (we will look at these more in II.B.2). We
have seen inverses before when working with exponents and radicals back in I.A.3 and I.A.5. They
are used to undo things when solving, and that’s what logarithms are essentially.
What is a log? A log is an exponent. That’s it. That’s the mystery. “log 2 ” literally translates to
“The exponent of 2 that results in 2 .” What’s the exponent of 2? 3! To be fair, log problems aren’t
always this simple, but we will develop tools to solve all exponentials and logs.
What is a logarithmic function? A logarithmic function is the inverse of an exponential function. In
other words a logarithmic function undoes an exponential function and an exponential function
undoes a logarithmic function. That’s the sole purpose of logs…to solve exponentials. So why make
it so complicated? The world may never know.
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One of the properties most students learn after much rearranging is the following:
log b = x
There is no reason to learn this after rearranging because this is what a log means. This is what it
was developed for. What’s the exponent of b that results in b ? Um…x. This is a property? Really?!
Example 1
Solve 4 = 1024 for x in terms of a log.
The base is 4, so we will use log to solve this problem. Remember when solving, whatever we do
to one side, we have to do to both sides.
4 = 1024
log 4 = log 1024
x = log 1024
Note: This is the same concept as “rearranging” but there is no need to memorize where to move
everything. Just solve the problem!
Now for some devastating news: solving for log 1024 does absolutely no good because there is no
log key on the calculator. So even though we can “rearrange” this much easier than before, it’s
completely pointless in solving for a numerical value. Angry yet?
There are two log keys on the calculator. “LOG” is equivalent to log , and we refer to it as the
standard log. “LN” is equivalent to log , and we refer to it as the natural log.
Note: The number e is called the natural number. It is an irrational number (cannot be written as a
fraction) that has many interesting properties that are investigated further in more advanced
mathematical courses. For our purpose, it’s about 2.7.
Example 2
Solve 10 = 1234 for x.
With a base of 10, we will use log or log, which means we can use our calculator.
10 = 1234 → log(10 ) = log 1234 → x = log 1234 = 3.091
Example 3
Solve e = 7 for x.
With a base of e, we will use log or ln, which means we can use our calculator to solve for x.
e = 7 → ln(e ) = ln 7 → 3x + 2 = ln 7 → x =
= −.018
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These are extremely helpful if we have a base of 10 or e, but out of the infinite number of numbers,
our odds of getting a problem with 10 or e are slim. So how do we use these to solve an exponential
with a different base? We are going to develop a property of logs that will allow us to use “log” or “ln”.
In I.A.3, we proved (b ) = b . In other words, to raise one exponent to another exponent, we
multiply the exponents. As a quick reminder, a log is an exponent. So when we take the log
(exponent) of an exponential expression, we can simply multiply the exponents together.
Example 1 (Redux)
Solve 4 = 1024 for x.
log b = x log b
Since log is not a calculator function, we can use log instead.
log(4 ) = log(1024)
x log(4) = log(1024)
x =
= 5 (Be sure to close the parentheses after each log)
In Example 1, we got an answer of log 1024, and in the Redux, we got an answer of
these are equal to each other and equal to 5. In fact, this is a perfect example of the Change of
Base Formula.
log a =
. Both of
Although the change of base formula works, it’s easier to solve the problem like we did in the Redux.
Example 4
Solve (3 ) = 6315 for x.
log(3 ) = log(6315)
(x + 6) log 3 = log 6315
x + 6 =
x =
− 6 = 7.965
So, how do we solve a logarithmic function? Remember, logs and exponentials are inverses of one
another, so to solve a log, we use an exponential. This is a really simple property.
b = x
Translating this one into English is a little tougher, but it should make sense.
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Example 5
Solve log (4x + 2) = 2 for x.
Since the log has a base of 5, we will use an exponential with a base of 5 to cancel it out,
remembering to do identical operations to both sides of the equation.
5 () = 5
4x + 2 = 25
x =
For advanced exponential and log operations we have two more log properties to assist us that derive
from a = a a and a =
, or in other words, to multiply (or divide) two terms with the same
base, we add (or subtract) the exponents.
log(ab) = log a + log b
log a = log a − log b
b
Example 6
Solve 2 log x + log 3 = 5 for x.
Before solving this, we have to simplify the log expression into a single log operation.
2 log x + log 3 = 5 → log x + log 3 = 5 → log (3x ) = 5
3 = 3
3x = 243
x = 81
x = ±9
Example 7
Solve 5(3 ) = 2000 for x.
log 5(3 ) = log 2000 → log 5 + (2x + 7) log 3 = log 2000
2x + 7 =
x =
= −0.773
Since an exponential with a positive base can never yield a negative result or zero, we get:
3 rd COMMANDMENT: Thou shalt only log positive numbers.
Final Note: log 1 = 0, and log b = 1.
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II.A.2 Independent Practice
Solve the following for x.
1. 3 = 243
11. ln x = 5
2. 5 = 125
12. log 4x = 2
3. 7 = 49
13. log (7x + 2) = 4
4. 2 + 5 = 37
14. ln(5x + 2) = 7
5. 9 + 8 = 89
15. log (8x ) = 3
6. 3(4 − 5) = 177
16. log (2x + 5) = 3
7. 5(2 − 4) + 2 = 22
17. ln(2x) + ln 5 = 2
8. 2 − 8(2 ) = 0
18. log (3x) + log x + = 3
9. 3e − 10e + 8 = 0
19. log (x + 5) + log (2x − 3) = 2
10. 7(3 ) + 12(3 ) − 4 = 0
20. log (2x + 5) = 6
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Section
II.A.3 Function
Division
In this section, we will divide one function by another function to get a quotient and a remainder.
Before we venture any further into function division, we need to remind ourselves what exactly a
remainder is and how to find it. We dealt with remainders in I.A.2, but we wanted our remainder to be
0 to convert a fraction (division problem) into its decimal equivalent. With functions, there is no such
thing as a decimal equivalent, so the remainder doesn’t just disappear. For instance, if we divide 23
by 3, we think “3 doesn’t go into 23 evenly, but it does go into 21 seven times”. The remainder here
is 2. Therefore, = + = 7 + . Mixed numbers like 7 are just plain awful. Never write your
numbers like this. Notice that the remainder is over what we divided by. This is always the case.
Synthetic Division
Synthetic Division is exactly what its name describes; it’s fake division. It allows us to divide two
polynomials without actually ever using division. We will only use multiplication and addition in
synthetic division. There are three rules for synthetic division.
1. Use zeros to represent any missing terms in the numerator.
2. The denominator must be of the form x + a, where a is any real number.
3. The quotient is always one degree less than the numerator.
Let’s look at the process for dividing
. There are no missing terms and the denominator
satisfies rule 2.
First we write out our grid:
2 3 5 −2
The 2 is the “zero” of the denominator. The 3, 5, and -2 are the coefficients of the
2 3 5 −2
2 3 5 −2
numerator. ↓ The 3 drops down as is. ↓ 6 The 3 multiplies by the 2 to get
3
2 3 5 −2
3
2 3 5 −2
6, which is placed in the next column. ↓ 6 We add 5 and 6 to get 11. ↓ 6 22
3 11
3 11 20
We multiply 11 by the 2 to get 22, which goes in the next column, and then we add -2 and 22 to get
20. We’re done with the calculations, but what does it all mean? Remember rule 3, the quotient is
always one degree less than the numerator. Since the numerator had degree of 2 (x ), the quotient
will have a degree of 1 (x). Therefore our answer is 3x + 11, but what is that 20? That’s the
remainder. Therefore the answer to this division problem is 3x + 11 +
This process is the same no matter how many terms there are in the numerator so no need to be
frightened.
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Example 1
Divide
.
The “zero” of the denominator is −3, and we’re missing the x term from the numerator. Thus we get
−3 5 0 2 −3 4
the initial grid:
Filling the grid out with the pattern from before we get:
−3 5 0
↓ −15
5 −15
Example 2
2 −3 4
45 141 414
47 138 418
Therefore the answer is 5x − 15x + 47x + 138 +
.
Divide
.
We can easily see that the x term is missing, but what may not be so obvious is the x-coefficient in
the denominator. What do we do? We have to get rid of the 2. So we divide both the numerator and
denominator by 2 (like reducing a fraction). We get
. Our grid starts as
1/2 3 0 −2 1
and ends as
So the answer is 3x + x − x +
1/2 3 0
↓ 3/2
3 3/2
−2
3/4
−5/4
or 3x + x − x +
1
−5/8
3/8
original denominator. Notice we doubled both the numerator and denominator.
The Remainder Theorem
if need the remainder to be over the
The Remainder Theorem is absolutely wonderful for calculating function values of unruly polynomials.
Essentially the remainder theorem states that if f(x) is a polynomial, f(a) is equal to the remainder of
()
.
Example 3
If f(x) = 5x − 2x + 3, then f(3) =
We synthetically divide
, and calculate the remainder.
The remainder is 120, and thus f(3) = 120.
3 5 −2
↓ 15
5 13
0
39
39
3
117
120
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II.A.3 Independent Practice
Divide the following using synthetic division.
1.
=
2.
=
3.
=
4.
=
5.
=
6.
=
7.
=
8.
=
Use the Remainder Theorem to solve for the following function values.
9. f(x) = 8x + 12x − 2. f(10) =
10. f(x) = 7x + 5x − 4x + 2. f(6) =
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Section
II.A.4 Polynomial
Functions
Polynomial functions should be nothing new to us. We’ve worked with both lines and parabolas.
Higher degree polynomial functions are simply extensions of what we have already learned. The
most interesting aspect of polynomial functions are their graphs, and, yet, graphing polynomials is
huge pain in the neck. In fact, if we really care about graphing, we can graph in a graphing calculator
or on a computer. What’s really important is the conceptualization behind the graphs. Can we look at
a polynomial function and know about what it will look like? Can we find the x-intercepts and/or zeros
of the function? By the end of this section we should be able to.
End Behavior
The End Behavior of a function has to do with what’s happening as it goes off forever in both
directions. Polynomials have only two options when it comes to their end behaviors: up or down.
And the end behaviors are determined by the highest exponent and the leading coefficient (discussed
in I.C.5). Why is this? As x approaches an infinitely big value, the highest degreed monomial
overpowers all other terms, so it’s the only one that matters. Sure, there might be curves and
craziness toward the middle, but we really do not care when we’re looking at the “ends”.
Odd-degreed polynomials, as we get away from the center and towards infinity (and negative infinity),
behave very much like a line. Therefore if the leading coefficient is positive, the polynomial will have
a positive “slope”, and if it is negative, it will have a negative “slope”. In other words the end behavior
of a positive odd-degreed polynomial is down to the left and up to the right. A negative odd-degreed
polynomial will be reversed (up-left and down-right).
Even degreed polynomials, as we approach the “ends”, behave very much like a parabola. So, if the
leading coefficient is positive, then the graph is concave up, and if negative, it is concave down
(discussed in I.C.4). The end behavior would then be described as up-left and up-right for a positive
leading coefficient, and down-left and down-right for a negative one.
Example 1
Determine the end behavior of f(x) = 3x + 4x − 2x .
Again, if we cared what the graph actually looked like, we could use technology. We only care about
what’s going on to the extreme left and extreme right. First we must reorder this polynomial in
descending order according to the degrees of the individual terms. We get f(x) = −2x + 4x + 3x.
The degree is 7, and the leading coefficient is negative. Therefore, the end behaviors look like a line
with a negative slope. The end behaviors are up-left and down-right.
Example 2
Determine the end behavior of f(x) = 2x + 6x + 8x .
In descending order we get f(x) = 8x + 2x + 6x . An even degreed polynomial with a positive
leading coefficient is concave up and has end-behaviors of up-left and up-right.
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Zeros and x-intercepts
End behaviors describe shape towards infinity; zeros and x-intercepts define shape everywhere inbetween.
Zeros are the values of x that yield a function value of 0. X-intercepts are the graphical
equivalent (where the graph intersects the x-axis). There is one significant difference between x-
intercepts and zeros. A polynomial function must have zeros but might not have x-intercepts. How is
this possible?
A function of degree n must have n zeros. However, there can be duplicates and imaginary zeros.
All x-intercepts must be real. So if there are no real zeros, there will be no x-intercepts. There’s also
the possibility of a double zero, where two zeros have the same value and appear as a single x-
intercept.
Single zeros appear as x-intercepts where the graph goes through the x-axis. Double zeros appear
as x-intercepts where the graph touches the x-axis and “bounces off”. Imaginary zeros do not appear
on the graph at all. We’re going to concentrate on finding x-intercepts (real zeros). We shouldn’t
worry ourselves too much with numbers that don’t actually exist. We’ll leave that to math nerds.
Finding x-intercepts
We’ve already spent time finding x-intercepts for lines (I.B.3) and parabolas (I.C.4), but what about
higher degree polynomials. After all, there is always a possibility of having up to 8 x-intercepts for an
eighth-degree polynomial. Finding these algebraically would be a ridiculous experience and a
completely pointless waste of our time. Granted many people would say that about most of
mathematics, but we’ll ignore them.
We’ll use a combination of calculator, synthetic division, and factoring (or quadratic formula) to find
the x-intercepts of a function.
Calculator Interlude
If we have a polynomial function with a degree of 3 or higher, we will use a graphing calculator to find
all but 2 of the x-intercepts, specifically integer x-intercepts. We go to “Y =” and input the function into
Y and 0 into Y . Then we press “GRAPH”. If there are no x-intercepts, then we’re done. If there are
we go to “2 ND ” “TRACE” 5 “ENTER” “ENTER”. At this point we’re asked to guess. But
isn’t the calculator supposed to just tell us? If we went through all this just to have to make the
decision ourselves, then why spend money on a calculator? The term “guess” isn’t the best way to
put it. The calculator can only find one intersection at a time, so it asks the user to decide which
intersection to find. We can use the left or right arrow keys to move the cursor to the intersection we
want the calculator to find and hit “ENTER”, or we can actually input a numerical guess and hit
“ENTER”. The calculator will then find the intersection closest to our “guess”. We will do this as
many times as it takes to find all but two of the intersection points. After that, we use our algebraic
skills we’ve been building.
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Example 3
Find the x-intercepts of f(x) = 10x − 31x + x + 6.
First thing we do is input the equation into Y and 0 into Y .
When we hit graph, we see three x-intercepts. The only clear integer value appears
to be 3. Let’s verify.
We now synthetically divide out 3 from our original polynomial.
3 10 −31 1 6
↓ 30 −3 −6 This leaves us with 10x − x − 2.
10 −1 −2 0
Factoring this we get (5x + 2)(2x − 1) = 0, which means the x-intercepts are 3, , and − .
Note: The last step may not be factorable. We will then need to use the quadratic formula,
x = ±()()
()
= ±
=
or − , which are equivalent to the answers we found by factoring.
This space has been left blank to allow us to breathe.
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II.A.4 Independent Practice
Arrange the following polynomials in descending order and describe their end behaviors.
1. y = 5x + 3x − 2
4. y = 3 − 5x − 6x
2. y = 6 − 4x + 3x − 2x
5. y = 5x + 7x − 2x + 3x
3. y = 9x + 5x + 2x − 1
6. y = 5 − 7x
Determine all x-intercepts of the following polynomials algebraically.
7. y = x − 7x + 12
12. y = 25x + 20x + 4
8. y = x + 6x − 16
13. y = 6x + 13x + 5x
9. y = 2x + 9x + 9
14. y = 4x + 32x + 85x + 75
10. y = 4x − 5x − 6
15. y = x − 4x + x + 6
11. y = 9x − 16
16. y = −4x − 8x + 51x − 45
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Section
II.A.5 Rational
Functions
If you remember, rational numbers are numbers that can be written as fractions. Likewise, rational
functions are functions that are written as fractions. In fact, a rational function is a ratio of two
polynomial functions. Just like polynomial functions, rational functions have x-intercepts and end
behaviors, but there are a few significant differences.
x-intercepts
The x-intercepts of a rational function occur when the numerator equals zero. Why? Because zero
divided by anything equals zero. Therefore, if the numerator of a function equals zero, then the
function itself equals zero. Since the numerator is a polynomial, we can solve for x-intercepts in the
same way we did in II.A.4.
Vertical Asymptotes
A vertical asymptote is a vertical line that acts like a boundary that the function’s graph cannot cross.
They form at an x-value because that x-value is not in the domain of the function. As a reminder, the
first commandment of mathematics says we cannot divide by zero. Therefore, any x-value that
results in a zero in the denominator cannot be in the domain. So we draw vertical dotted lines at
these x-values to show the graph doesn’t contain these values.
Holes
Holes are weird. Keep in mind, since a rational function is a big, fat fraction, there is the possibility of
factors cancelling out. When this happens, we get a simpler fraction. However, any value that results
in zero in the denominator (even if it cancels out) still cannot be in the domain. Holes appear in the
graph because they are individual points that have been removed because the x-value is not allowed
in the domain of the function. In other words, holes occur at x-values that result in a zero in both the
numerator and denominator.
Example 1
Find the x-intercepts, vertical asymptotes, and holes of f(x) = ()()
()() .
The numerator and denominator are already factored for us, so this should be relatively simple.
Setting the numerator equal to zero, we get x-values of − and 2. Setting the denominator equal to
zero, we get x-values of −3 and 2. Since there is a 2 for both, there is a hole at x = 2. We will find it
in a second. There is an x-intercept at − , 0 and a vertical asymptote at x = −3. If we cancel out
x − 2 from the numerator and denominator, we end up with
. Plugging 2 in for x we end up with
. Therefore, we have a hole at 2, .
As you can see the processes for finding these three values is all the same. We just have to
remember which is which.
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Horizontal and Slant Asymptotes
Even though horizontal and slant asymptotes share a name with vertical asymptotes, they are vastly
different. Vertical asymptotes are a boundary, a wall that cannot be crossed. Horizontal and slant
asymptotes form a guideline for end behavior. They can be crossed multiple times as long as the end
behaviors approach them.
When the degree of the numerator is less than the degree of the denominator, we have a horizontal
asymptote at y = 0, regardless of the coefficients and actual degrees. For instance f(x) =
has
a horizontal asymptote at y = 0 because the degree of the numerator is 1 and the degree of the
denominator is 2. Why does this happen? Remember, when we talk about end behavior, we’re
talking about infinity and at infinity only the largest values matter. Therefore, we only look at the
largest parts of this function, namely
. If this was our fraction, an x would cancel out, leaving . If
we allow x to get larger the fraction would get smaller, and if we let x get infinitely big, the fraction
would get infinitely small, so small in fact, that it might as well be zero.
When the degree of the numerator is equal to the degree of the denominator, we have a horizontal
asymptote at y = where a is the leading coefficient of the numerator, and b is the leading coefficient
of the denominator. For instance f(x) =
has a horizontal asymptote at y = because the
degrees are equal and is the ratio of the leading coefficients. This happens for the same reason
illustrated above. When dealing with large values, all that matters is
, and x cancels out, leaving
.
When the degree of the numerator is one more than the degree of the denominator we get a slant
asymptote that is equal to the quotient after dividing the function. For example, f(x) =
has a
slant asymptote at y = 4x − 3. We get this through synthetic division:
−2 4 5 −2
↓ −8 6 The remainder of
doesn’t matter because for infinitely large values of x, the
4 −3 4
remainder equals zero. Additionally, if we were to focus on the parts that matter at large x-values,
, cancel out x to get 4x, we see that we have a 1-degree polynomial with a positive leading
coefficient. Therefore, we know the end behavior has to be down-left and up-right. The line,
y = 4x − 3 shares this end behavior.
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II.A.5 Independent Practice
Determine the x-intercepts, vertical asymptotes, horizontal or slant asymptotes, and holes of
the following rational functions.
1. y =
2. y =
3. y =
4. y =
5. y =
6. y =
7. y =
8. y =
9. y =
10. y =
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Chapter
II.B
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Chapter
II.B Chapter
Overview
‣ II.B.1 – Composite Functions
This section explores composite or embedded functions with numerical, variable, and functional
inputs.
‣ II.B.2 – Inverse Functions
This section will distinguish inverse function from reciprocal function notation. It will also discuss
the relationship between a function and its inverse, explain the significance of invertibility, and
allow for the algebraic identification of a function’s inverse.
‣ II.B.3 – External Transformations
This section will transform graphs through vertical translation and dilation by manipulation of the
function externally.
‣ II.B.4 – Internal Transformations
This section will transform graphs through horizontal translation and dilation by manipulation of the
function internally.
‣ II.B.5 – Transformations Involving Absolute Value
This section will investigate the effects of absolute value on the graph of a function, both internally
and externally. It will also combine all transformations into single examples.
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Section
II.B.1 Composite
Functions
Composite functions are functions made up of inner and outer functions. For instance
f(x) = (x + 5) has an inner function x + 5 and an outer function x . The inner function is much
easier to understand than the outer function. We actually see x + 5 in f(x), so it makes sense.
Where does x come from? Since x is a variable, it can actually represent any value or, in this case,
any expression. So when we embed these two functions the inner function, x + 5, replaces x in the
outer function, x , yielding (x + 5) . Now let’s take a few steps back.
Composite Function Notation
If we have to functions, f and g, and we want to embed g inside of f, we write this composite function
as either fg(x) or (f°g)(x). Both of these are read “f of g of x”, however, the first notation is much
more intuitive.
So let’s say f(x) = 3x + 2 and g(x) = 2x − 5, and we are asked to find fg(3). According to Order
of Operations we work inside out, so we find g(3). g(3) = 1. Now we can replace g(3) with 1, and
we get f(1). f(1) = 5. This is our final output, so fg(3) = 5.
Example 1
If f(x) = 3x + 2 and g(x) = √x, then fg(16) =
Firstly we find g(16) = 4, and then we find f(4) = 14. Therefore fg(16) = 14.
Creating a Composite Function
Up until this point, we’ve only substituted a single value in for x, but more often we have to plug
multiple inputs into fg(x). Plugging each value into one function and then plugging in the result
into another function can get tedious after a while. We need to be able to compile a single function in
order to make this process as easy as possible.
Let’s say f(x) = 3x + 2 and g(x) = 5x − 1. We want to find fg(x). In this problem x is the input
and the output is 5x − 1, which becomes the input for f. We now need to find f(5x − 1), which
means we substitute 5x − 1 in for x in the f function.
f(5x − 1) = 3(5x − 1) + 2 = 15x − 3 + 2 = 15x − 1.
Now that we have one simple function, we can plug in as many numerical values as we desire without
breaking a sweat.
Example 2
If f(x) = 3x + 5 and g(x) = 4 , find fg(x).
fg(x) = f(4 ) = 3(4 ) + 5
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Expression Inputs
What happens if the initial input is an expression? What if we’re asked to find fg(x + 2)? What do
we do? We treat x + 2 like an inside function the same way we had. In a lot of ways, it would be
easier to find fg(x) first and then find fg(x + 2).
Example 3
Given f(x) = 3x + 7 and g(x) = 2x − 1, fg(x − 1) =
Firstly let’s find fg(x). fg(x) = f(2x − 1) = 3(2x − 1) + 7 = 6x − 3 + 7 = 6x + 4.
So we have found fg(x) = 6x + 4. If we want to find fg(x − 1), all we have to do is substitute
x − 1 in to the composite function. fg(x − 1) = 6(x − 1) + 4 = 6x − 6 + 4 = 6x − 2. Therefore,
fg(x − 1) = 6x − 2.
Example 4
Given f(x) = 3 and g(x) = log x, find fg(x) and gf(x).
fg(x) = 3 = x and gf(x) = log 3 = x They both equal the same thing, and the input is the
same as the output. This is significant but not always true. In fact, it’s rarely true. We’ll investigate
this more in the next section.
Example 5
Given f(x) = 3x + 2 and g(x) = x , find fg(x) and gf(x).
fg(x) = 3(x ) + 2 = 3x + 2 and gf(x) = (3x + 2) = (3x + 2)(3x + 2) = 9x + 12x + 4. These
are not nearly the same thing, nor does the input match the output in either case. Example 4 is
significant.
Calculator Interlude
In I.B.1, we investigated function notation in the calculator. We can also use this same functionality to
perform operations with composite functions.
Finding a Numerical Answer
Input f(x) into Y and g(x) into Y . Then press “2 ND ” “QUIT”. Once on the home screen, we’ll input
Y Y (a), where a is the numerical input. (Reminder: Go to “VARS” “Y-VARS” “1: Function” to
get to Y and Y ) Here we do Example 1 with the calculator.
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II.B.1 Independent Practice
Given f(x) = 3x + 2 and g(x) = 2x 2 .
1. fg(2) =
4. gf(x) =
2. gf(−3) =
5. fg(2x) =
3. fg(x) =
6. gf(x + 2) =
Given f(x) = e x and g(x) = ln(2x).
7. fg(3) =
10. gf(x) =
8. gf(−2) =
11. fg(x ) =
9. fg(x) =
12. gf(x − 2) =
Given f(x) = √x + 5 and g(x) = x 2 − 5.
13. fg(−5) =
16. gf(x) =
14. gf(4) =
17. fg(x − 3) =
15. fg(x) =
18. gf(e ) =
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Section
II.B.2 Inverse
Functions
In the last section, we noted that Example 4 was significant. The reason why it’s significant is
because f(x) = 3 and g(x) = log x are inverses. We discussed this fact in II.A.2. Remember
log b = x and b = x because they cancel one another out. This leads to the following definition
of inverse functions:
If f and g are inverse functions of one another, fg(x) = gf(x) = x. Additionally, we say
g(x) = f (x) and f(x) = g (x), where f (x) is read as either “the inverse of f of x” or “f inverse of
x”.
One quick note: the −1 may appear to be the exponent that indicates the reciprocal of a value. A
reciprocal is not an inverse. Here’s the difference: f (x) is the inverse function and f(x) is the
reciprocal function. When a −1 is the superscript of the function name, it indicates inverse.
Otherwise, it indicates reciprocal.
Example 1
Are f(x) = 3x + 2 and g(x) = 2x − 1 inverses of one another?
fg(x) = f(2x − 1) = 3(2x − 1) + 2 = 6x − 3 + 2 = 6x − 1
Since fg(x) ≠ x, g(x) ≠ f (x).
Example 2
Are f(x) = 2x + 5 and g(x) = x − inverses of one another?
fg(x) = f 1 2 x − 5 2 = 2 1 2 x − 5 + 5 = x − 5 + 5 = x
2
gf(x) = g(2x + 5) = 1 2 (2x + 5) − 5 2 = x + 5 2 − 5 2 = x
Since fg(x) = gf(x) = x, g(x) = f (x).
Example 3
Are f(x) = x and g(x) = √x inverses of one another?
fg(x) = f√x = √x = x
gf(x) = g(x ) = x = |x|
Since gf(x) ≠ x, these are not inverse functions of one another. However, they are inverse
operations. This leads us into our next topic.
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Invertibility
A function is invertible if its inverse is also a function. Remember, to determine if an equation is a
function, it must pass the vertical line test (I.B.1). The inverse of a function must pass its own vertical
line test to prove the inverse is a function as well. So how do we graph an inverse?
The graph of the inverse of a function is a reflection of the graph of the function over the line y = x. If
we look at the graphs of f(x) = 3 and g(x) = log x, which we already know to be inverses of one
another, we get:
As we can see the two graphs are reflections of one another over y = x.
Additionally, we see that both f and g pass the vertical line test, therefore both
f and g are functions. This does seem like a lot of work, however, to
determine if a function is invertible, especially when the inverse is unknown.
An easier method is the horizontal line test. Exactly like the vertical line test,
a function passes the horizontal line test if there are no horizontal lines that
can be drawn that intersect the function more than once. If a function passes the horizontal line test,
it is invertible.
Calculating Inverses
Let’s have a look at the graphs of Example 3. As we can see f(x) = x fails
the horizontal line test, so its true inverse (a horizontal parabola, y = ±√x) is
not a function. On the other hand, g(x) = √x does pass the horizontal line
test, so its inverse is a function. However, the true inverse of √x is only the
right half of a parabola with a restricted domain of x ≥ 0. Graphically, it makes
sense that f and g are not inverses of one another.
Yes, we can determine if an inverse will, in fact, be a function, but we also need to determine an
inverse, algebraically. The process in a nutshell is (a) switch x and y, and (b) solve for y. The
solving bit can get a bit tricky.
Example 4
Determine the inverse of f(x) = 3x + 4.
Because f is a line and passes the horizontal line test, its inverse is also a function. First we switch x
and y, x = 3y + 4, and then solve for y. 3y = x − 4 → y =
→ y = x − . Because the inverse is
a function, f (x) = x − .
Example 5
Determine the inverse of f(x) = √x + 5 − 2.
x = y + 5 − 2 → x + 2 = y + 5 → (x + 2) = y + 5 → x + 4x + 4 = y + 5 → y = x + 4x − 1.
f (x) = x + 4x − 1.
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II.B.2 Independent Practice
Find the inverses of the following functions.
1. f(x) = 3x + 5
2. g(x) = 4x − 8
3. h(x) = √x + 2
4. p(x) = log (x − 7)
5. q(x) = 8
6. r(x) = ln(5x + 7)
7. s(x) = 5
8. t(x) = e
9. v(x) = 5(x + 9)
10. b(x) = 3x + 12x + 5
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Section
II.B.3 External
Transformations
In the next three sections we’re going to investigate the effects that modifying aspects of a function
has on its graph. In this one, we’re going to look at vertical movement and stretching.
Vertical Translation
Let’s begin with a quadratic function, f(x) = x − 3, and a transformation of it, g(x) = f(x) + 2 =
x − 1. How do the two graphs compare. Remember, to graph a quadratic equation, we simply
plugged in several points. Here are two tables showing the outcomes for both functions.
f(x)
x
-2 1
-1 -2
0 -3
1 -2
2 1
As we can see, the y-values of g are 2 higher than each of the corresponding y-values of f. This
means that by adding 2 to f, the entire graph is moved up 2. This should make sense since we are
adding a value to f(x), a y-value itself. Let’s have a look at the graphs.
g(x)
x
-2 3
-1 0
0 -1
1 0
2 3
The good news is that no matter what type of function f(x) may be, these same rules apply. If we
add a number to the outside of a function, the graph translates vertically by that amount. If we add a
10, then the graph has a vertical translation of 10 (up 10). If we add a -4, then the graph has a
vertical translation of -4 (down 4).
Example 1
Describe the graphical transformation from f(x) = x + 5x − 2 to g(x) = x + 5x + 7.
Since the equation essentially stays the same except for the constant, we can actually view g(x) as
being equivalent to f(x) + 9, since 7 is 9 more than -2. Therefore, the entire graph has moved up 5.
We should refer to this as a vertical translation of 5.
Example 2
Find the equation of f(x) = 2 + 6 after it has been vertically translated by -5.
The new equation will be of the form g(x) = f(x) − 5 = 2 + 6 − 5 = 2 + 1.
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Vertical Dilation
Often, in algebra classes, teachers will tell their students that the a in ax + bx + c controls the width
of the parabola. If a > 1, they say the parabola is “thin”. If a < 1, they say the parabola is “wide”. If
a = 1, they say the parabola is “normal”. These are all lies. The width does not change, and we’ll
prove it.
Let’s look at tables for f(x) = x − 4 and g(x) = 2f(x) = 2(x − 4) = 2x − 8.
f(x)
x
-3 5
-2 0
-1 -3
0 -4
1 -3
2 0
3 5
3 10
If we inspect these next to one another, we can see that the g(x) values are twice the f(x) values
(since that’s how it was defined). The points of g(x) themselves are twice as far from the x-axis as
the corresponding points of f(x). The graph isn’t getting thinner, it’s stretching vertically. For proof,
look at the x-intercepts (-2,0) and (2,0). They are the same for both functions. If the graph did get
thinner, wouldn’t these be closer together in g(x)? The graph of g(x) may, indeed, look thinner, but
that’s merely a byproduct of the vertical stretching. When you stretch a rubber band vertically,
doesn’t it get “thinner” too? The stretching or dilation is the cause, the relative “width” is the effect.
Here is the graph of the two functions.
g(x)
x
-3 10
-2 0
-1 -6
0 -8
1 -6
2 0
Example 3
Describe the graphical transformation from f(x) = log x to g(x) = 5 log x.
The graph is stretched vertically, such that each y-value is multiplied by 5. We say the graph is
vertically dilated by a scale factor of 5.
Example 4
Determine a function that begins as f(x) = √4x + 2 − 8, is dilated vertically by a scale factor of , and
is translated vertically by 6.
g(x) = 1 2 √4x + 2 − 8 + 6 = 1 2 √4x + 2 − 4 + 6 = 1 2 √4x + 2 + 2
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Vertical Reflection
When we say a graph reflects vertically, we mean it reflects across the x-axis (y = 0). The x-axis is a
mirror, so to speak. So let’s say we want to reflect the point (3,5) across the x-axis. It’s currently a
distance of 5 above the x-axis, and so we want it to go to a point that is 5 below the x-axis. We want
to move (3,5) to (3,-5). It would appear we want to move the point down 10. However, we would
want to move a point (6,-2) to (6,2), which is a translation of 4, vertically. How can we accomplish
both of these movements with one operation? We need to negate the y-values of course!
Therefore g(x) = −f(x) is a vertical reflection of f(x).
Example 5
Determine an equation for g(x) that vertically reflects f(x) = 3x + 2x − 5.
g(x) = −f(x) = −(3x + 2x − 5) = −3x − 2x + 5.
Order of Operations
Order of operations takes a significant role in transformations. If multiple transformations take place
in the same problem, in what order to they transform the original function? Remember GEMA (I.A.3),
says Grouping, Exponents, Multiplication, Addition. Since no transformations include changing an
exponent, we can ignore the E in GEMA.
Example 6
What transformations take place from f(x) = 3x to g(x) = −24x + 7?
g(x) = −8f(x) + 7. In order, we have a vertical reflection, vertical dilation by a scale factor of 8, and
a vertical translation of 7.
Example 7
What transformations take place from f(x) = 3x to g(x) = 7(3x + 5)?
We have two options: g(x) = 7(f(x) + 5), which is a vertical translation of 5 and a vertical dilation by
a scale factor of 7, or g(x) = 7f(x) + 35, which is a vertical dilation by a scale factor of 7 and a
vertical translation of 35. Both of these result in the same graph.
Terminology
Depending on the textbook, the names of f(x), the original graph, and g(x), the transformed graph
differ greatly. In some, they are known as the pre-image and the image, respectively. In others, they
are known as the image and after-image. This text will refer to them by the second set. The
reasoning: If no transformation takes place, we just have an image. Only when we transform it, do
we create a second or after image. “Pre-image” dictates that a transformation must take place, and
that’s not always the case.
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II.B.3 Independent Practice
Describe the transformation that occurs from f to g.
1. g(x) = 3f(x)
4. g(x) = −f(x) + 6
2. g(x) = f(x) + 4
5. g(x) = f(x) − 5
3. g(x) = 2f(x) − 3
6. g(x) = −3f(x) + 2
Describe the transformation that occurs from f to g.
7. f(x) = √x → g(x) = 3√x − 5
10. f(x) = 2 → g(x) = −2 − 1
8. f(x) = e → g(x) = −2e − 3
11. f(x) = 2x + 1 → g(x) = 4x + 2
9. f(x) = ln x → g(x) = ln x − 9
12. f(x) = 4x + 8 → g(x) = −2x + 7
Write a function, g, that describes the transformation from the thin graph, f, to the thick graph.
13.
y
g
f
x
15.
y
g
f
x
14.
y
x
f
g
16.
y
f
x
g
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Section
II.B.4 Internal
Transformations
In this section, we will look at modifications inside a function and their effects on the graph of the
function. One thing to keep in mind: all internal transformations act in the opposite way of what we
should expect, especially now that we know all about external transformations.
Horizontal Translation
When adding a value to the outside of a function, we were changing the y-value. So, if we modify the
inside of a function by adding values, we are changing the x-value, and, thus, we should expect
horizontal translation (left and right movement).
Let’s look at f(x) = x − 3 and g(x) = f(x + 2) = (x + 2) − 3 = x + 4x + 1 in terms of a table of
values. Before we do, however, we would probably assume, since we’re adding 2 to the x-value, the
graph will move right 2. Let’s see if our assumption is true.
If g(x) were a simple horizontal translation of 2, then the vertex would move from (0,-3) to (2,-3), but
we see in the table that (0,-3) actually moved to (-2,-3), the opposite of our assumption. What’s going
on? f(x) is still the function we’re dealing with, but the x-values are changing. Each x-value is
having 2 added to it before being inputted into f, which means that the y-value usually attributed to an
x-value is actually being attributed to an x-value that’s less by 2. For instance, f(1) = −2, however
the 1 is actually a -1 that has had 2 added to it. Therefore (1,-2) would be a point on the graph of f,
but (-1,2) would be the corresponding point on the graph of g. This graphically creates a leftward
translation.
Example 1
Describe the transformation from f(x) = (x − 4) to g(x) = (x + 7) .
Essentially g(x) = f(x + 11), which indicates a horizontal translation of -11.
Example 2
x f(x)
-2 1
-1 -2
0 -3
1 -2
2 1
x g(x)
-2 -3
-1 -2
0 1
1 6
2 13
Determine an equation for g that transforms that translates the graph of f(x) = 4x − 4x + 1 right 2.
One option is
g(x) = f(x − 2) = 4(x − 2) − 4(x − 2) + 1 = 4(x − 4x + 4) − 4(x − 2) + 1 = 4x − 20x + 25.
The other option is first realizing f(x) = (2x − 1) .
So g(x) = f(x − 2) = (2(x − 2) − 1) = (2x − 5) = 4x − 20x + 25.
We get the same answer either way, and both shift the graph right 2.
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Horizontal Dilation
Since multiplying a scalar value by the y-value results in vertical dilation, it should make sense that
multiplying a scalar value by the x-value would result in horizontal dilation. For example, if we let
f(x) = x − 4 and g(x) = f(2x) = (2x) − 4 = 4x − 4, we may expect that the function would be
horizontally stretched to twice its width. Let’s look at the values.
If the graph were to stretch horizontally by a scale factor of 2, we would expect that (-1,-3) would
become (-2,-3), but what actually appears to be happening is that a point like (-2,0) has become
(-1,0), where -1 is actually of -2. Again this is the opposite of what we expected to happen. The
reasoning for this is similar to the reasoning for horizontal translation. In g, each x-value that goes
into f is actually twice the actual x-value. Therefore every x-value is receiving the y-value it would get
if the x-value were twice as big. Normally, -2 would go in and 0 would come out, but g is bulking up
its x-values, producing 0 from an initial input of -1. Nifty, huh?
Essentially, when we multiply the x-value in a function by a scalar multiple, the actual x-values on the
graph are divided by that same scalar value. In other words, f(ax) would dilate f(x) horizontally, by
a scale factor of , the reciprocal of a.
Example 3
x f(x)
-4 12
-2 0
-1 -3
0 -4
1 -3
2 0
4 12
x g(x)
-4 60
-2 12
-1 0
0 -4
1 0
2 12
4 60
Describe the transformation from f(x) = 3 ln x to g(x) = 3ln ( x).
g(x) = f x which indicates a horizontal dilation by a scale factor of 2.
Example 4
Determine an equation for g(x) that dilates f(x) = 7 sin(x + 5) horizontally by a scale factor of .
Here, we’re presented with a strange function, but it doesn’t matter. It works the same way as all
functions. Therefore g(x) = f(4x) = 7 sin4(x + 5) = 7sin (4x + 20)
Note: Typically we would leave our answer as 7 sin4(x + 5) as this is standard form, but very often
we will be presented with 7sin (4x + 20), and it’s not immediate as to what the actual transformation
is supposed to be.
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Horizontal Reflection
Horizontal translation and horizontal dilation both acted in the opposite manner of what was expected.
With reflection, there really isn’t a way to reflect in an opposite manner, so reflection is immune to the
opposite-ness of the previous two concepts. With vertical reflection across the x-axis, we negated the
y-values. Likewise, for horizontal reflection across the y-axis (x = 0), we negate the x-values.
So g(x) = f(−x) reflects a graph horizontally.
Example 5
Describe the transformation from f(x) = 3tan (x − 5) to g(x) = 3 tan(−x − 5).
g(x) = 3 tan(−x − 5) = f(−x). f undergoes a horizontal reflection across the y-axis.
Order of Operations
Order of operations is a bit tricky when dealing with inner transformations. If we let f(x) = 3x and
g(x) = 3(2x − 6) , we would expect to have a horizontal dilation by a scale factor of and a
horizontal translation right 6. This is a false assumption. Remember, everything inside a function
happens in the opposite manner of what is expected. This applies to GEMA as well. Therefore the
translation (addition) happens before the dilation (multiplication), but in this example, the -6 is affected
by the dilation term (2). Therefore, we have to factor out the 2. g(x) = 32(x − 3) . Now we can
see that there is a horizontal translation of 3 and a horizontal dilation by a scale factor of .
Example 6
Describe the transformations from f(x) = log (−6x) to g(x) = log (−3x + 8).
g(x) = f x + 8 = f (x + 16), which shoes a horizontal translation of -16 and a horizontal
dilation by a scale factor of 2.
Example 7
Create a function g(x) that will transform f(x) = 3√2x + 6 + 5 by a horizontal translation of 3, a
horizontal dilation of , and a horizontal reflection.
g(x) = f−3(x − 3) = 32−3(x − 3) + 6 + 5 = 3√−6x + 24 + 5
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II.B.4 Independent Practice
Describe the transformation that occurs from f to g.
1. g(x) = f(3x)
4. g(x) = f(2x + 6)
2. g(x) = f(x + 5)
5. g(x) = f x − 2
3. g(x) = f − x
6. g(x) = f(−3x − 9)
Describe the transformation that occurs from f to g.
7. f(x) = x → g(x) = (x + 5)
10. f(x) = log x → g(x) = log (−2x + 10)
8. f(x) = √x → g(x) = x
11. f(x) = √−x + 7
→ g(x) = x − 4
9. f(x) = 2 → g(x) = 2
12. f(x) = sin(x + 6) → g(x) = sin(−2x − 4)
Write a function, g, that describes the transformation from the thin graph, f, to the thick graph.
y
y
g
x
x
f
f
g
13.
15.
y
y
x
f
g
x
f
g
14.
16.
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Section
II.B.5 Transformations
Involving Absolute Value
We dealt with absolute value back in I.A.5 when we were solving for all values of x in an equation. In
this section we’re going to see how absolute value affects a graph, both externally and internally. The
order of operations is extremely important when it comes to absolute value. Not only are they
considered a form of grouping, they perform an operation as well. In a second, we will see why they
get their very own section.
External Absolute Value
To begin, let’s review exactly what absolute value does. It determines a numbers positive distance
from zero. In other words it has no effect on positive numbers, but it negates negative numbers to
make them positive. Therefore if we have g(x) = |f(x)|, we will leave the positive values of f(x)
alone and negate the negative values, and if memory serves correctly, negating the y-value reflects it
across the x-axis. Therefore an external absolute value will vertically reflect all points (and by
extension all parts of the graph) below the x-axis across the x-axis.
Example 1
Describe the transformation from f(x) = 3x + 2 to g(x) = |3x + 2|.
Since g(x) = |f(x)|, then all points below the x-axis will be reflected above the x-axis as shown by
the graphs of f and g.
Example 2
Describe the transformation from f(x) = 4x − 6 to g(x) = |4x − 4|.
g(x) = |4x − 6 + 2| = |f(x) + 2|. Therefore we have a vertical translation of 2 and then a reflection
of all points below the x-axis across the x-axis.
Example 3
Describe the transformations from f(x) = sin (2x) to g(x) = −5|sin (2x)| − 2.
g(x) = −5|f(x)| − 2. We have a vertical reflection of all points below the x-axis across the x-axis,
then a vertical reflection of the entire graph across the x-axis, then a vertical dilation by a scale factor
of 5, and finally a vertical translation by -2.
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Internal Absolute Value
Once again the internal transformations still have an unexpected outcome. In this case, we’re leaving
positive x-values alone and negating the negative x-values. This means that all negative values will
now take on the same y-values as their positive counterparts, but nothing on the right side of the y-
axis moves. In essence what happens is that everything to the left of the y-axis becomes a mirror
reflection of everything to the right of the y-axis. All order of operations still hold true, and we still
have to factor out any scalar multiples.
Example 4
Describe the transformation from f(x) = e to g(x) = e || .
Here, g(x) = f(|x|). Therefore all points to the left of the y-axis reflect those to the right.
Example 5
Describe the transformation from f(x) = cos(x) to g(x) = cos(2|x| + 4).
g(x) = f(2|x| + 4) = f2(|x| + 2). All points to the left of the y-axis reflect those to the right, then
the graph horizontally translates -2, and then the graph horizontally dilates by a scale factor of .
Example 6
Describe the transformation from f(x) = x to g(x) = 5(|−x + 5|) − 4.
g(x) = 5f(|−x + 5|) − 4 = 5f(|−(x − 5)|) − 4. Working outside in we have a horizontal reflection, a
horizontal translation of 5, a mirror reflection of all positive x-values, a vertical dilation by a scale
factor of 5, and a vertical translation of -4.
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II.B.5 Independent Practice
Describe the transformations from f to g.
1. g(x) = 2f(4x)
5. g(x) = 2|f(4x)| + 3
2. g(x) = f(x + 3) − 4
6. g(x) = |−5f(5x + 10) − 4|
3. g(x) = 2f(2x + 6) + 4
7. g(x) = f(|x|) + 9
4. g(x) = −4f x − 3 + 2
8. g(x) = 3f x − 5 − 1
Describe the transformations from f to g.
9. f(x) = 3 → g(x) = 2(3 ) − 5
13. f(x) = √x → g(x) = 33|x| + 9 − 2
10. f(x) = x → g(x) = 5(2x + 8) + 9
14. f(x) = e → g(x) = |5e − 7|
11. f(x) = ln x → g(x) = ln x + 1 + 7
15. f(x) = x → g(x) = −3(x + 2) − 5
12. f(x) = sin x → g(x) = 5 sin(6x − 12) − 3
16. f(x) = √x
→ g(x) =
−3|x| + 6
+ 7
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Chapter
II.C
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Chapter
II.C Chapter
Overview
‣ II.C.1 – Linear Systems Graphically
This section will examine the graphical representations of a system and its solution by use of a
calculator’s “Intersection” feature.
‣ II.C.2 – Elimination Method
This section will investigate the use of the elimination method to solve a system of standard form
linear equations.
‣ II.C.3 – Substitution Method
This section will utilize substitution to solve a system of linear equations regardless of form.
‣ II.C.4 – Non-Linear Systems
This section will prepare students to solve systems of linear equations, quadratic equations, power
equations, and exponential equations.
‣ II.C.5 – System Applications
This section will utilize the knowledge gained in the previous four sections to solve word problems.
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Section
II.C.1
Linear Systems
Graphically
A system of equations is simply a set of equations whose solution is the point of intersection of all of
the equations. The majority of the time, the systems of equations will consist of two equations with
two unknowns. Any more than that, and we will have to use matrices to solve the system (II.D.4).
In this section, we will solve a system of two linear equations by graphing, and we will do so using the
Intersect feature in the calculator as we did in II.A.4. The good news is that because two lines can
only have a maximum of one intersection, we won’t have to “guess” this time.
To refresh: Plug the first equation into Y and the second equation into Y . Press GRAPH 2ND
TRACE 5 ENTER ENTER ENTER. That’s it. The calculator will give you both the x and
y-values of the point of intersection.
Example 1
Find the solution of
y = 4x + 2
y = −x + 12 . The solution is (2,10).
Example 2
3x − y = 5
Find the solution of
y = −2x + 6 .
Firstly, we have to solve the first equation for y. y = 3x − 5.
Now because the solution is not a pair of integers, hit 2ND QUIT. Once on the home screen, hit
ALPHA X MATH Frac ENTER and ALPHA Y MATH Frac ENTER.
The solution is , .
Example 3
y = 3x + 2
Solve
y = 3x − 5 .
Solving this we get
, which means there is “No Solution” because the lines are parallel.
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II.C.1 Independent Practice
Graph the following systems, and solve.
y 2x
5
1.
y 3x
10
2.
y
y
3
4
3
2
x 2
x 6
y
y
x
x
3.
y
1
x 4
2
x 3y
3
y
2x
5y
10
4.
2
y x 3
5
y
x
x
5. 4 x 2 y 3
8x
4y
6
6. 3 x 3 y 6
x 2y
10
y
y
x
x
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Section
II.C.2 Elimination
Method
The Elimination Method is a method by which we can solve systems of linear equations algebraically.
However, it is essential that both linear equations be of the same form.
Example 1
3x + 4y = 18
Solve the system
5x − 4y = −2 .
If we add the two equations together, we end up with 8x = 16, which can then be solved for x = 2.
We plug 2 in for x in either of the equations 3(2) + 4y = 18 → 4y = 12 → y = 3. To check our answer
we plug our solution into the second equation 5(2) − 4(3) = 10 − 12 = −2. Thus, the solution is
(2,3).
Example 2
3x + 5y = −7
Solve the system
2x − 10y = 22 .
Adding these two equations together doesn’t eliminate either variable, but if we multiply the first
6x + 10y = −14
equation by 2, we get . Now adding these together leaves us with 8x = 8 → x = 1.
2x − 10y = 22
Then 6(1) + 10y = −14 → 10y = −20 → y = −2. The solution is (1,-2).
Example 3
2x − 4y = 5
Solve the system
−3x + 6y = 7 .
This time, we have to multiply both equations by different scalars to cancel something out. We will
cancel out the x’s by multiplying the first equation by 3 and the second equation by 2. This will yield
6x − 12y = 15
. Adding the two equations together, both x and y cancel out, leaving 0 = 29.
−6x + 12y = 14
Since we have a false statement, there is No Solution.
Example 4
6x − 4y = 20
Solve the system
y = .
x − 5
Firstly, we’ll have to change the second equation to standard form by first multiplying the bottom
6x − 4y = 20
equation by 2 and getting x and y on the same side. We end up with . Doubling
−3x + 2y = −10
6x − 4y = 20
the bottom equation gives us , and adding these together yields 0 = 0. This is a
−6x + 4y = −20
true statement, therefore both lines are, in fact, the same line. There are infinitely many solutions.
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II.C.2 Independent Practice
Solve the following systems with elimination.
3x
2 y 5
1.
4x
2 y 9
6.
y 2x
7
4x
2 y 3
2.
1
2
x 7y
3
x 4y
0
1
2
7.
2x
5y
8
4x
7 y 18
3.
y 2x
4
y 5x
5
8.
5x
3y
17
10x
5y
10
4.
3x
y 2
6x
2y
4
9.
2x
3y
9
3x
4y
22
5.
2x
2y
18
2x
5y
24
11x
7 y 2
10. 3x
2 y 19
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Section
II.C.3 Substitution
Method
The Substitution Method is a much more versatile method for solving a system algebraically. In some
cases it’s easier than the elimination method, and in other cases, elimination is easier. Elimination is
easier when we are given two standard form linear equations. Substitution is easier the rest of the
time. See? It is versatile. Plus, we can use substitution with non-linear systems, which we will see in
the next section.
Example 1
y = 3x + 2
Solve the system
y = 2x + 6 .
Because both equations equal the same variable, we can set the two equations equal to one another.
3x + 2 = 2x + 6. We can then solve for x. x = 4. Just like with elimination we plug 4 in for x, and we
get y = 3(4) + 2 = 14. The solution is (4,14).
Example 2
3x + 2y = 4
Solve the system
x = 6y − 2 .
Since we have one equation solved for x, we will substitute the expression in for x in the other
equation: 3(6y − 2) + 2y = 4 → 20y = 10 → y = . We can substitute this into either equation, but
since we’re looking for x, it would be prudent to plug it into the second equation: x = 6 − 2 = 1.
The solution is 1, .
Example 3
4x − 2y = 8
Solve the system
y = 2x − 4 .
We plug the second equation into the first equation: 4x − 2(2x − 4) = 8 → 4x − 4x + 8 = 8 → 8 = 8.
Just like in elimination, if both variables have been cancelled, and the statement is true, the equations
graph the same line, and there are infinitely many solutions (every point on the line is in fact a
solution).
Example 4
x = 3y − 5
Solve the system
2x = 6y + 3 .
This may appear to be an elimination type problem, but since one of the equations is solved for x,
we’ll use substitution. Plugging the first equation in to the second equation yields
2(3y − 5) = 6y + 3 → 6y − 10 = 6y + 3 → −10 = 3. Since this gives a false equation, there is No
Solution.
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II.C.3 Independent Practice
Solve the following systems of equations with substitution.
1.
7
2
3
2
y
x
x
y
2.
8
2
6
2
1
y
x
y
x
3.
5
2
2
3
x
y
x
y
4.
6
2
3
5
2
y
x
y
x
5.
36
3
12
12
4
y
x
y
x
6.
5
4
8
x
y
x
y
7.
2
11
5
3
x
y
x
8.
12
5
10
7
2
y
x
y
x
9.
7
2
3
4
y
x
y
10.
6
2
21
7
6
y
x
y
x

Section
II.C.4 Non-Linear
Systems
There are four other types of systems of equations that can be solved algebraically (with just a little
help from the calculator).
Example 1 (A Linear Equation and a Quadratic Equation)
y = 3x + 5
Solve
y = 2x − 5x + 13 .
Because of the nature of a line and a parabola, there can be no solution, one solution, or two
solutions. We can see that the two equations are both set equal to y, so we can set them equal to
each other. 3x + 5 = 2x − 5x + 13 → 0 = 2x − 8x + 8. Using the quadratic formula:
x = ±()()
()
=
Example 2 (Two Quadratic Euqations)
Solve y = 2x + 5x − 3
y = −4x + 8x + 42 .
= 2 and y = 3(2) + 5 = 11. The solution is (2,11).
Setting the equations equal to one another: 2x + 5x − 3 = −4x + 8x + 42 → 0 = 6x − 3x − 45.
Now we use the quadratic formula: x = ±()()
= ±
= 3 or − . We have to plug both of
()
these into one of the equations to get their y-coordinates. y = 2(3) + 5(3) − 3 = 30 and
y = 2 − + 5 − − 3 = −3. Our two solutions are (3,30) and − , −3.
Example 3 (Two Exponential Functions)
y = 2(4)
Solve
y = (8).
To solve this, we will use a modified elimination method, where we divide the second equation by the
first. We get 1 = (2) → 8 = 2 → ln 8 = x ln 2 → x =
= 3. Then y = 2(4) = 128. The solution
is (3,128).
Example 4 (Two Power Functions)
y = 3x
Solve
y = x.
To solve this, we will divide the second equation by the first: 1 = x → x = 4 and y = 3(4) = 48.
The solution is (4,48).
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II.C.4 Independent Practice
Solve the following Non-Linear Systems of Equations.
1. y = 4x + 6
y = 3x − 2x − 39
5. y = 5x
y = 20x
2. x − 3y = −7
y = 2x + 3x − 11
6. y = 4x
y = 36x
3. y = 5x + 6x − 4
y = −4x + 8x + 3
7. y = 5(4)
y = 40(2)
4. y = −3x − 5x + 7
y = 2x − 6x − 11
8. y = 4050
y = 800
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Section
II.C.5 System
Applications
In this section we’ll use our knowledge of systems to solve the dreaded word problems. Chilling…
II.C.5 Independent Practice
Answer the following questions.
1. Walter went to the store for some prunes and Metamucil. Don’t ask me why. He bought three
jars of prunes and four packages of Metamucil. Overall he spent $29.
a. Write a standard form equation of this situation, allowing x to be the price of prunes and
y to be the price of Metamucil.
b. Convert the equation to slope-intercept form.
2. Janice went to the same store to purchase 5 jars of prunes and 2 packages of Metamucil.
Overall she spent $25.
a. Write a standard form equation of this situation, allowing x to be the price of prunes and
y to be the price of Metamucil.
b. Convert this equation to slope-intercept form.
c. What is the price of prunes?
d. What is the price of Metamucil?
3. A farmer sells both chicken and duck eggs at the local farmer’s market. If he sells 10 dozen
duck eggs and 20 dozen chicken eggs, he will make $87.50. If he sells 15 duck eggs and 25
dozen chicken eggs, he will make $118.75. How much do a dozen eggs of each species cost?
4. Cher has $200 in her savings account. She is able to save $25 a week from her paycheck.
She also owes $500 to her credit card, and she is paying off $50 a week. At what point will
Cher have the same amount of money in her savings as what she still owes in credit?
5. Dayv needs to make a backdrop. He buys 7 buckets of bargain paint and 5 buckets of good
paint. He spends $175. Later on he buys 4 buckets of bargain paint and 2 buckets of good
paint. He spends $68. How much does a bucket of each cost?
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Chapter
II.D
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Chapter
II.D Chapter
Overview
‣ II.D.1 – Matrix Addition & Scalar Multiplication
This section will lay the ground work for adding and subtracting matrices, as well as multiplying a
matrix by a scalar multiple. It will detail the difference between rows and columns with an
emphasis on the dimensions of a matrix.
‣ II.D.2 – Matrix Multiplication
This section will employ the concept of multiplying two matrices together, discussing when it is
and when it is not prudent to do so, performing the operation by hand and by calculator.
‣ II.D.3 – Determinants & Inverse Matrices
This section will allow for the determination of determinants and inverse matrices by hand and
with a calculator.
‣ II.D.4 – Systems of Equations
This section will introduce a matrix method for solving polynomial systems known as Reduced
Row Echelon.
‣ II.D.5 – Matrix Transformation
This section will develop the applicable concepts of matrix mathematics and its geometric and
graphical links.
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Section
II.D.1
Matrix Addition &
Scalar Multiplication
What is the Matrix? It’s a movie starring Keanu Reeves. What is a matrix? It’s a multidimensional
representation with many advanced applications. Before we can get to some of those
applications, we need to learn the basics of matrices.
Matrices 101
A matrix takes on the form of a grid inside brackets: X = x x x
x x x
. This is a 2 × 3 matrix
because it has 2 rows and 3 columns, and the elements x are in the i row and j column.
y
y
Y = is a 3 × 1 matrix, and Z = [z
y
number of rows and columns (i.e. x
Matrix Addition
x
x
z ] is a 1 × 2 matrix. Additionally, a matrix with the same
x
) is considered a square matrix.
To add two matrices together, they must have the same dimensions, i × j. Then, we add their
corresponding values: x + y , x + y , etc…
Example 1
3 5 −2
4 7 9 + 2 −1 7
−4 3 1 =
3 + 2 5 − 1 −2 + 7
4 − 4 7 + 3 9 + 1 = 5 4 5
0 10 10
Scalar Multiplication
Having discussed scalar multiples in the previous section, this should be a piece of cake. Essentially,
scalar multiples transform all values by dilating each value by the scale value. Functionally, we
multiply every value in a matrix by the scalar multiple. The process is very similar to distribution.
Example 2
4 −5
3 7 1 =
2 10
3(4) 3(−5) 12 −15
3(7) 3(1) = 21 3
3(2) 3(10) 6 30
That’s it. Those are the basics…
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II.D.1 Independent Practice
Simplify the following.
1. 3 5
2 9 + 5 2
7 9 =
2. 7 −2
0 5 + −5 1
9 −7 =
3. 6 2 −5
3 7 −8 − 4 7 9
−4 −7 −3 =
3 0 5
4. 2 −6 4
−3 8 −2
−3 2 8
+
12 −4 −2
9 −5 10
=
5 8 3 7
5. 2 4 − 3 −2 4 =
−7 6 −5 −6
6. 5 3 −4
6 2 + 6 4 5
−7 5 =
7. 3 2 9 4 2
4 −1 −7 + 4 2 7 =
−4 8
8. 2[5 12 −8] − 9[5 9 10] =
9. 6 5 −3 3 −4
+ 10
4 −6 −8 3 =
6 −6 5
10. 7 −2 − 3 5 + 8 2 =
5 2 −4
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Section
II.D.2 Matrix
Multiplication
Matrix multiplication is distinctly different from matrix addition. In fact, they are not similar at all.
Many students will say 3 5
2 4 5 2 = 15
10
, using the same technique as addition. They would
5 1 10 4
be so very, very wrong.
General Guidelines
1. Matrix multiplication is NOT commutative. AB ≠ BA.
2. The number of columns of the first matrix must match the number of rows of the second
matrix. (i.e. i × j and j × k)
3. The matrix product will have the same number rows as the first matrix and the same number of
columns of the same matrix. ((i × j)(j × k) = i × k)
4. To get x of the product matrix, we must get the summed product of row i of the first matrix
and column j of the second matrix.
Example 1
2 4 1 2
3 5 6 7 =
1
The dimensions of the first matrix and second matrix are 2 × 3 and 3 × 1, respectively. Because the
number of columns in the first matrix match the number of rows in the second matrix, we can multiply
them together. The product matrix will have dimensions 2 × 1. All we have to do is multiply the
elements in the first row (of the first matrix) by the corresponding elements in the first column (of the
second matrix) and add their products together, we then repeat with the second row and first column.
2(2) + 4(7) + 1(1) + 28 + 1
= 4
3(2) + 5(7) + 6(1) 6 + 35 + 6 = 33 47
Example 2
2 −3
4 1 −1 3 8
3 −2 1 =
The dimensions 2 × 2 and 2 × 3 imply that the product will have dimensions 2 × 3.
2(−1) − 3(3) 2(3) − 3(−2) 2(8) − 3(1)
12 13
= −11
4(−1) + 1(3) 4(3) + 1(−2) 4(8) + 1(1) −1 10 33
Example 3
2 3 6
1 5 8 [ 3 2 1] =
The dimensions of the first, 2 × 3, and the second, 1 × 3, make it impossible to multiply these
together. There is No Solution.
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II.D.2 Independent Practice
Simplify the following.
1. 4 5
8 3 2 7
4 8 =
4
2. [5 6 − 2] 7 =
−3
3. 5 6 2 −4
−2 7 9 5 =
2
6 9
4. 2 − 2 5 − 1
3 2 =
−3 5
5. 3 5
2 6 4 5 2
7 9 1 =
5 1
6. 7 2 2 6 3
4 1 8 =
3 1
−2
7. 5 [3 − 2 5] =
2
8. 4 1 − 2
4 8 3 5 − 1
−2 8 =
5 2 1 5
9. 7 4 2 −2 =
9 6 2 −1
5 2 − 3 3 − 1 7
10. 6 − 1 2 −2 7 3 =
−7 4 8 5 8 − 3
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Section
II.D.3 Determinants &
Inverse Matrices
There is no such thing as matrix division, but we can perform a similar operation by multiplying by an
inverse matrix. To define an inverse matrix, we need to first define an identity matrix. An identity
matrix is a square matrix that takes the appearance of 1 0 1 0 0
0 1 and 0 1 0, where the main
0 0 1
diagonal contains 1s and all other spaces are 0. If M is a matrix and I is the identity matrix,
MI = IM = M. It’s like multiplying a number by 1.
A square matrix M has an inverse, M , if MM = M M = I.
Example 1
Determine if 3 5
2 1 and −
−
are inverses of one another.
3 5
2 1 −
− = 3 − + 5
3 + 5 −
2 − + 1
2 + 1 − = − + −
− +
− = 1 0
0 1
These are indeed inverses of one another. Technically, we would need to switch them around and
multiply again, but we’re not going to because we’re already freaking out here.
Determinants
In order to calculate an inverse matrix, we first need to calculate the determinant. The determinant of
a b
c d is written with absolute value marks a b , which is defined as
c d
a
c
b = ad − bc
d
If the determinant of a matrix equals zero, it is non-invertible.
Example 2
3 5
6 2 =
3(2) − 5(6) = 6 − 30 = −24
To calculate the determinant of a 3 × 3 matrix, we have to calculate three separate 2 × 2
determinants. To be fair, we won’t be doing this by hand much, but the concept will reappear when
we talking about three-dimensional vectors (IV.C.4).
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a b c
If we’re asked to find d e f, we draw a line through the first row and first column, multiplying a by
g h i
a b c
the resulting 2 × 2 matrix: d e f → a(ei − fh). Moving the line to the second column, we repeat
g h i
a b c
the process: d e f → b(di − fg). Moving the line to the third column, we repeat the process one
g h i
a b c
more time: d e f → c(dh − eg). We then add the first determinant, subtract the second, and add
g h i
the third: a(ei − fh) − b(di − fg) + c(dh − eg).
Example 3
3 2 1
5 3 2 =
5 1 4
33(4) − 2(1) − 25(4) − 2(5) + 15(1) − 3(5) = 3(10) − 2(10) + 1(−10) = 30 − 20 − 10 = 0
This matrix has no inverse because the determinant equals zero.
Determining an Inverse Matrix
The process of finding an inverse matrix is actually quite simple. If M = a b
c d ,
M = 1
|M| d
−c
−b
a
Example 4
2 5
3 1 =
Firstly, we find the determinant: 2(1) − 5(3) = 2 − 15 = −13. We then use the formula for inverse
matrices: − 1
−3
−5
2 = −
−
.
Finding the inverse of a 3 × 3 matrix is way too complicated to do it by hand (unless we get bored
with everything else life has to offer, so we’ll use a calculator for any matrices larger than 2 × 2.
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Calculator Interlude
If we go to the matrix menu (2ND x ) and scroll through the page titles using the right arrow key,
we get the following screens.
The NAMES screen allows us to take a matrix we’ve created and use it on the home screen for
calculations. The MATH screen allows to find the determinant (1:det() and other information we’ll get
to in later sections. The EDIT screen allows us to create and define matrices. Let’s go to the EDIT
screen and choose 1:[A]. We get the following screen:
In this screen we can change the dimensions and input the data from our matrix problem. Once we
input everything we press 2ND → QUIT. Now we can find the determinant by going back to the
matrix menu on the MATH screen or we can use the x key to calculate the inverse, using the matrix
names from the NAMES screen.
Example 5
3 2 1 3 2 1
Calculate 5 6 2 and 5 6 2
7 8 3 7 8 3
and then determine if the answer is, in fact, correct.
First, we need to input the matrix into [A].
Then we calculate the determinant and inverse matrix.
To check out answer, we have to show that AA = A A = I.
1 1 −1
It does, so the determinant is 2, and the inverse matrix is −
1 − .
−1 −5 4
Yes, we can do matrix operations in the calculator, but there are times when doing matrix operations
by hand will be necessary, so it’s good to go ahead and practice using pencil and paper to do the
easier problems.
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II.D.3 Independent Practice
Calculate the following by hand.
1. 6 2
5 4 =
4. 4 5
−2 − 2 =
2. 1 9
3 6 =
5. −5 − 2
3 − 7 =
3. −2 4
6 − 1 =
6. 4 9
−12 − 27 =
7. 2 8
3 1 =
9. −2 6
4 − 12 =
8. 8 − 2
5 − 3 =
10. 5 2
8 0 =
Calculate the following with a calculator.
4 5 2
11. 7 4 1 =
3 5 9
5 3 4
13. 3 9 1
9 4 7
=
2 − 5 5 6
12.
6 7 − 1 9
=
−2 5 9 8
9 4 3 − 6
5 − 2 6 5
14.
−1 9 2 9
8 7 − 3 10
2 9 8 − 7
=
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Section
II.D.4
Systems of
Equations
One application that matrices are ideal for is solving systems of equations, and there are actually two
matrix methods for solving a system, the Inverse Matrix Method and the Reduced Row Echelon
Method. They both have the same guidelines for getting started:
1. If a system has n different variables, then the system must have n different equations.
2. Both methods only work with polynomials in standard form (i.e. Ax + By + Cz = D)
Inverse Matrix Method
The inverse matrix method uses two general concepts: [A B
x
C] y = Ax + By + Cz and
z
M MA = IA = A.
Example 1
3x + 2y + 5z = 22
Solve 2x − 4y + 3z = 3 .
4x + y − 2z = 0
3 2 5 x 22
We can recreate this as 2 −4 3 y = 3 . If we multiply both sides by the inverse matrix, we
4 1 −2 z 0
3 2 5 3 2 5 x 3 2 5 22
get 2 −4 3 2 −4 3 y = 2 −4 3 3 . The inverse matrices cancel out, and
4 1 −2 4 1 −2 z 4 1 −2 0
x 3 2 5 22
we end up with y = 2 −4 3 3 . Now we use the calculator.
z 4 1 −2 0
x 1
Since y = 2, the point of intersection of the system is (1,2,3).
z 3
Reduced Row Echelon Form (rref)
The explanation for rref is really, really easy to ignore, so we will. rref can be found in the MATRIX
MATH screen down at letter B. For this method we will create a single matrix out of the standard
form system and have the calculator solve the system for us. Once it does, we will end up with a
1 0 0 x
matrix like 0 1 0 y where (x,y,z) is our answer. Notice the identity matrix making another
0 0 1 z
appearance. One big advantage this method has is that sometimes an inverse matrix cannot be
found (determinant = 0), and so the inverse matrix method just won’t work.
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Example 2
3x + 2y + 5z = 22
Solve Solve 2x − 4y + 3z = 3 .
4x + y − 2z = 0
Let’s create a 3 × 4 matrix:
3 2 5 22
2 −4 3 3
4 1 −2 0
We plug this into the calculator. For this the fourth column is off-screen on the far right. We can
scroll over to it.
Again, we see the same solution we found in Example 1, (1,2,3).
No Solution and Infinitely Many Solutions
Remember no solutions occurs when there is no point of intersection of the three equations (II.C),
and infinitely many solutions occurs when the three equations overlap in a line (or a plane), when
there are infinitely many points of intersection. When using the rref method, the calculator will say
1 0 a
0 1 c
0 0 0
Example 3
b
1 0 a
d when there are infinitely many solutions and 0 1 c
0
0 0 0
3x + 2y − 6z = −3
Solve 4x − 5y + 2z = 3 .
6x + 4y − 12z = −7
b
d when there are no solutions.
1
Because the last row has a 1 at the end of it, there are no solutions.
What if we could tell the system was not solvable without having to do all the calculator input. Well,
there is a way. The general rule is if one row is a multiple of another row, there are infinitely many
solutions. If one row is a multiple of another row except for the last value, there are no solutions.
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II.D.4 Independent Practice
Solve the following systems of equations.
1. 3x + 4y = 18
5x − 2y = 4
4. 3x + 2y = 14
−6x − 4y = −30
2. 2x − 5y = −3
4x + 3y = 7
5. 7x − y = 9
x + 6y = 32
3. 6x + 2y = 8
5x − 5y = 20
6. 8x + 2y = 30
10x + y = 33
7. 5x + 2y − 7z = 0
3x + 6y + 2z = 11
4x − y + 6z = 9
9. 3w + 6x − 2y + z = 2
5w − 4x − 8y + 9z = 98
2w + 5x + 4y − 8z = −63
−3w + 2x − y + 2z = 1
8. 3x − 5y − 2z = 8
4x − 8y + 6z = −28
−5x + 2y + 7z = −47
10. 5w − 4x + 7y − 2z = 68
6w + 3x − 8y − z = −52
7w − 5x + 4y + 8z = −23
18w + 9x − 24y − 3z = −156
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Section
II.D.5 Matrix
Transformation
Another application of matrices is transformations of geometric polygons through dilation, reflection,
rotation, and translation. To begin, an n-gon (a polygon with n sides and n interior angles) can be
represented by a 2 × n matrix, where each column represents a vertex of the polygon, making sure to
order the points in adjacent order. For the purpose of this section let quadrilateral Q be represented
by Q = 2 5 1
−3 4 0
Dilations
2
−1 .
We’ve actually already dealt with dilations in section II.D.1. In order to dilate a figure by a scale
factor, we multiply the image matrix by a scalar multiple. For instance, if we dilate a figure by a scale
factor of 2, then each side-length of the figure will double in length, and the distance from each point
to the origin, (0,0), will also double in length. For dilations with scale factors less than one, the figure
will shrink in size and move closer to the origin.
Example 1
Dilate Q by a scale factor of 5.
5 2 5 1
−3 4 0
Translations
2
=
10 25 5
−1 −15 20 0
10
−5
Translations were also introduced in II.D.1 as well with the addition of two matrices. To translate an
image matrix of a triangle right a and down b, we would add the triangle matrix to the matrix
a a a
, that way each x-value moves right a and each y-value moves down b.
−b −b −b
Example 2
Translate Q left 5 and up 3.
−5 −5 −5
3 3 3
−5
3 + 2 5 1
−3 4 0
2
0 −4
= −3
−1 0 7 3
−3
2 .
Reflections and Rotations
These will take some memorization, but for the most part, it shouldn’t be to bad. Firstly, both
reflections and rotations utilize matrix multiplication, and all of the important reflections and rotations
look awfully similar to the identity matrix.
Additionally, it’s important to know that a full rotation has 360° in it, and, unless stated otherwise,
rotations are in a counter-clockwise direction. Additionally, 90° counter-clockwise is equivalent to
270° clockwise (they add up to 360°), and 180° in either direction is the same.
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The following is a list of transformation matrices and their corresponding operation:
1 0 – Does Nothing
0 1
1 0 – Negates the y-values / Reflects vertically across y = 0.
0 −1
−1 0 - Negates the x-values / Reflects horizontally across x = 0.
0 1
0 1 – Exchanges the x-values and y-values / Reflects across y = x.
1 0
0 −1 - Exchanges and negates the x-values and y-values / Reflects across y = −x.
−1 0
0 1 - Rotates 90° counter-clockwise / 270° clockwise
−1 0
0 −1 - Rotates 270° counter-clockwise / 90° clockwise
1 0
−1 0 - Rotates 180°
0 −1
Example 3
Reflect Q across the x-axis and then rotate it 90° counter-clockwise.
To achieve this we will need to multiply 1 0
0 −1 0 1
−1 0 2 5 1 2
, and we can do this in the
−3 4 0 −1
calculator. The after-image is −3 4 0 −1
. As we can see, the x-values and y-values switched,
2 5 1 2
which means we could have achieved the same after-image by reflecting across y = x.
Example 4
Reflect Q across y = −x, dilate by a scale factor of 3, and translate right 2 and down 7.
3 0 −1
−1 0 2 5 1
−3 4 0
2
−1 + 2 2 2
−7 −7 −7
2 11 −10 2
=
−7 −13 −22 −10
5
−13 .
Example 5
Translate Q right 2 and down 7, dilate by a scale factor of 3, and then reflect across y = −x.
This is similar but opposite of Example 4, which means
3 0 −1
−1 0 2 5 1 2
−3 4 0 −1 + 2 2 2 2 30 9 21
=
−7 −7 −7 −7 −12 −21 −9
The order of transformation matters!
24
−12 .
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II.D.5 Independent Practice
Determine the vertices of the image after performing the following transformations to the
given pre-image:
1 2 5 7
3 6 8 4 .
1. Translate left 5 and up 6.
2. Dilate by a scale factor of 5.
3. Reflect across the x-axis.
4. Rotate 180°.
5. Translate right 3 and down 7. Then dilate by a scale factor of 2.
6. Reflect across y = x. Then rotate 90°.
7. Reflect across the y-axis. Then dilate by a scale factor of 8.
8. Rotate 270°. Then translate left 8 and down 2.
9. Translate right 5 and up 6. Dilate by a scale factor of 3. Reflect across y = −x. Rotate 180°.
10. Dilate by a scale factor of 7. Reflect across y = 0. Translate down 4 and right 8. Rotate 270°.
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Chapter
II.E
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Chapter
II.E Chapter
Overview
‣ II.E.1 – Recursive & Explicit Formulas
This section explores the concept of sequences through pattern recognition and generation, using
both a recursive and explicit approach.
‣ II.E.2 – Arithmetic & Geometric Sequences
This section formalizes the idea of arithmetic and geometric sequences, developing a standard
idea of explicit formula.
‣ II.E.3 – Arithmetic Series
This section introduces sigma notation as a method for demonstrating sums of arithmetic
sequences.
‣ II.E.4 – Geometric Series
This section investigates the partial sums and infinite series of geometric sequences.
‣ II.E.5 – Applications of Sequences & Series
This section utilizes the skills and concepts learned in previous sections to apply them to word
problems.
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Section
II.E.1
Recursive & Explicit
Formulas
In this section, we’re going to investigate numerical patterns and how to define them using formulas.
We will begin with something called Recursive Formulas.
Recursive Formulas
A recursive formula is one that relies on previous terms in the pattern, or sequence, to define current
terms. A term in a sequence is takes on the form t , where n is called the term number. If we define
the term as t , then the previous term is t , and the one before that is t , and so forth. A
recursive formula requires two things:
1. The definition of t (and sometimes t ).
2. The definition of t in terms of previous terms.
Example 1
Determine a recursive formula for the sequence 3,5,7,9,11,13, …
We see two things: (a) t = 3 and (b) 2 is being added to each term to get the next term, or
t = t + 2.
Example 2
Determine a recursive formula for the sequence 2,6,18,54,162, …
In this sequence, each term is 3 times the previous term. So t = 2 and t = 3t .
Example 3
Determine a recursive formula for the sequence 1,1,2,3,5,8,13, …
This is the famous Fibonacci sequence mentioned way back on page 8. Each term is the sum of the
previous two terms, which means we must define the first two terms. t = 1, t = 1, and
t = t + t .
Calculator Interlude
The graphing calculator has a recursive (or sequential) mode that allows for the definition of recursive
formulas, albeit in a different form. For instance in the calculator t is defined as u(n), v(n), or w(n).
Where u, v, and w allow for three different formulas to be used at once (u, v, and w can be found as
the second function of 7, 8, and 9 on the calculator. Do not use the alphabetic U, V, and W to define
formulas, as these are viewed as variables by the calculator. If we press MODE, we can see the
fourth line down has FUNC highlighted. We need to highlight SEQ by scrolling over and pressing
ENTER. Now when we press Y=, we see a completely different screen. Additionally, when pressing
the ‘X, T, θ, n’ button, we get an n instead of an X, now. This button changes function when the
calculator changes modes. So now that we’re somewhat familiar with this mode, let’s rework
Example 1.
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We get to the final screen by pressing 2ND GRAPH.
The following shows the Fibonacci sequence:
Explicit Formulas
The problem with recursive formulas is that to find the 50 th term, we need to know the 49 th term, but
then we need the 48 th term, and so on… This could take a while. We need a method that allows for
finding the 50 th term directly knowing simply that it is term 50. In other words, we need an explicit
formula that defines t in terms of n. Certain sequences, like the Fibonacci sequence, require
recursion, but the majority of the time, we would rather use explicit formulas.
Example 4
Determine an explicit formula for 3,5,7,9,11, …, and then calculate t .
We need to determine a rule for each term, t , in terms of its term number, n. In other words we
need a rule that takes in 1 and outputs 3 and also takes in 2 and outputs 5. The first thing we notice
is that 2 is being added to each term. This looks familiar. It looks like a linear pattern. Therefore, we
start with t = 2n + b. Using the first term, 3, we get t = 2(1) + b = 3. b = 1, and our explicit
pattern becomes t = 2n + 1, and t = 2(10) + 1 = 21.
Example 5
Determine an explicit formula for 2,6,18,54,162, …, and calculate t .
This appears to be an exponential pattern, where we’re multiplying by 3 each time. So we start with
t = a(3) . Using the first term, t = a(3) = 2, and a = . Therefore, t = (3) , and
t = (3) = 39366.
Example 6
Determine an explicit formula for , , ,
,
, …, and calculate t .
This pattern appears to be fractions made up of squares, specifically the term number’s square over
the next term number’s square. The explicit formula is t =
() , and, thus, t =
=
.
Note: Make sure we put our calculator back in FUNC mode once we finish calculating recursive
formulas.
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II.E.1 Independent Practice
Determine Recursive and Explicit formulas for the following patterns, and determine t 10 .
1. 3, 10, 17, 24, 31, …
6. 5, 10, 20, 40, 80, …
2. 5, 2, -1, -4, -7, …
7. 6, 18, 54, 162, 486, …
3. 7, -3, -13, -23, -33, …
8. 1, -4, 16, -64, 256, …
4. 8, 12, 16, 20, 24, …
9. -3, 15, -75, 375, -1875, …
5. 9, 16, 23, 30, 37, …
10. 64, 32, 16, 8, 4, …
Determine Explicit formulas for the following patterns, and determine t 10 .
11. 1, 4, 9, 16, 25, …
15. , , , , …
12. 3, 8, 15, 24, 35, …
16. , , , , , …
13. 8, 27, 64, 125, 216, …
17. , , , , , …
14. 6, 14, 24, 36, 50, …
18. , 2, , , , …
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Section
II.E.2
Arithmetic & Geometric
Sequences
We dealt with arithmetic and geometric sequences in the last section, but here we will formalize the
explicit definitions.
Arithmetic Sequences
Arithmetic sequences take on linear patterns that have a common difference, d, between each set of
subsequent terms. They have a recursive rule of t = t + d. The explicit form is
where t is any term we actually know.
Example 1
t = t + d(n − 1) or t = t + d(a − 1)
Determine an explicit formula for the sequence 5,8,11,14,17, …, and calculate t .
Since t = 5, and d = 3, t = 5 + 3(n − 1), and t = 5 + 3(10 − 1) = 32.
Example 2
Determine the term number, n, of the final term in the sequence 5,7,9,11,13, … ,65.
t = 5 + 2(n − 1) = 65, and solving this equation for n, we get get n = 31.
If all we’re given is two terms in the sequence t and t , we calculate d as
Example 3
d =
Given two terms t = 7 and t = 27 of an arithmetic sequence, determine the explicit formula for the
sequence.
The common difference in this case can be determined by d =
= 5. Therefore, the explicit
formula is t = 7(5) .
Geometric Sequences
Geometric sequences take on exponential patterns that have a common ratio, r, between each set of
subsequent terms. They have a recursive rule of t = rt and an explicit form of
t = t (r) or t = t (r) .
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Example 4
Determine an explicit formula for the sequence 2,10,50,250,1250, …, and calculate t .
Since t = 2, and r = 5, t = 2(5) , and t = 2(5) = 3906250.
Example 5
Determine the term number of the final term of the sequence 3,6,12,24,48, … ,98304.
t = 3(2) = 98304. Solving for n, we have to use logs.
2 = 32768
(n − 1) log 2 = log 32768
log 32768
n = + 1 = 16
log 2
To calculate the common ratio of a geometric sequence when give terms t and t ,
Example 6
r =
Determine the explicit formula for the geometric sequence with t = 8 and t = 2.
Finding r is a bit awkward in this problem, but not too awkward. r = = = . Therefore, the
explicit formula is t = 8 .
There’s not a whole lot more to it. We’ve already done all of this before. We’re just tying it all
together into a neat little bow.
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II.E.2 Independent Practice
Determine explicit formulas for the following Arithmetic sequences, determine t 10 , and
determine n for the final given term.
1. 5, 8, 11, 14, 17, … , 80
4. 24, 17, 10, 3, -4, … , -410
2. 7, 13, 19, 25, 31, … , 211
5. 72, 59, 46, 33, 20, … , -162
3. 12, 28, 44, 60, 76, … , 1404
6. 5, -4, -13, -22, -31, … , -1111
Determine explicit formulas for the following Geometric sequences, determine t 10 , and
determine n for the final given term.
7. 5, 10, 20, 40, 80, … , 40960
8. 7, -21, 63, -189, 567, …, -100442349
9. 4, 6, 9, 13.5, 20.25, …, 102.515625
10. 120, 60, 30, 15, 7.5, … , 0.05859375
11. 531441, 177147, 59049, 19683, 6561, …, 243
12. 1000000, 400000, 160000, 64000, 25600, … 1638.4
Determine both Arithmetic and Geometric formulas that contain the following pairs of terms.
13. t = 12, t = 6
17. t = 4, t = 16
14. t = 5, t = 15
18. t = 5, t = −135
15. t = 7, t = 21
19. t = 6, t = 150
16. t = 8, t = 12
20. t = 9, t = 72
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Section
II.E.3 Arithmetic
Series
An Arithmetic Series is the sum of an arithmetic sequence. We use something called sigma
notation to represent a partial arithmetic series.
t + d(n − 1)
This is read as “the sum from n = 1 to n = a of the arithmetic sequence t .”
Example 1
2 + 4(n − 1) =
2 + 4(0) + 2 + 4(1) + 2 + 4(2) + 2 + 4(3) + 2 + 4(4) = 2 + 6 + 10 + 14 + 18 = 50
This is an easy process as long as we have a small number of terms. So how do we calculate the
sum given a significantly larger set of terms? In the preceding example, if we average the first and
last terms, we get 10. Also, if we average the next inside terms, we get 10, and the middle term is
also 10. And if we multiply 10 by the number of terms, 5 in this case, we get the sum, 50. Nifty, huh?
There’s a story of a future mathematician who was disruptive in elementary school. His teacher sent
him to the corner, telling him he couldn’t rejoin the class until he had added all the integers from 1 to
100. He returned within a minute by using the concept above and the formula below.
t + d(n − a)
= t + t
2
In this formula b − a + 1 is the total number of terms, and
terms.
Example 2
4 + 5n =
∙ (b − a + 1)
is the average of the first and final
In this series, there are 20 − 3 + 1 = 18 terms. The first term is 4 + 5(3) = 19 and the last term is
4 + 5(20) = 104. The average of these two terms is
= 61.5. Therefore the sum is
61.5(18) = 1107.
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What happens if we’re not given a neat little formula, but instead we are given a partial arithmetic sum
of the individual terms?
Example 3
4 + 7 + 10 + 13 + 16 + ⋯ + 82 =
We know the first term, 4, and the last term, 82, which have an average of 43. But we do not know
how many terms there are. So, we use the skills we learned in II.E.2 and figure out t .
t = 4 + 3(n − 1) = 82. Therefore, n = 27. The sum of this arithmetic series is 43(27) = 1161.
That wasn’t so bad, but, what if we’re given the pattern and the sum, and we have to find how many
terms there are in the series?
Example 4
If ∑
(3 + 5n) = 2728, what is b?
We know that there are b − 2 + 1 = b − 1 terms. We also know that the first term is 13 and the last
term is 3 + 5b, which have an average of
= 8 + 2.5b. So the sum
2728 = (b − 1)(8 + 2.5b) = 2.5b + 5.5b − 8. Therefore, 2.5b + 5.5b − 2736 = 0, and using the
quadratic formula, we find either b = 32 or b = −34.2. Since a term number can only be a positive
integer, b = 32.
The good news is that we won’t be expected to do any of these on our own, but the process is an
important one to see.
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II.E.3 Independent Practice
Calculate the following Arithmetic Series.
1.
2.
3.
4.
5.
3n + 2 =
4n + 9 =
8n + 4 =
5n − 6 =
−3n + 2 =
6.
7.
8.
9.
10.
−4n + 12 =
3n + 8 =
4n − 5 =
3n + 2 =
n + 2 =
11. 5 + 8 + 11 + 14 + 17 + ⋯ + 65
14. 12 + 5 − 2 − 9 − 16 + ⋯ − 387
12. 7 + 15 + 23 + 31 + 39 + ⋯ + 367
15. 22 + 14 + 6 − 2 − 10 + ⋯ − 714
13. 9 + 14 + 19 + 24 + 29 + ⋯ + 164
16. 88 + 79 + 70 + 61 + 52 + ⋯ − 119
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Section
II.E.4 Geometric
Series
The geometric series has a more complicated formula. The good news is that all we need to
calculate a geometric series is the first term (t ), common ratio (r), and the number of terms
(b − a + 1).
t (r)
= t 1 − r
1 − r
Example 1
2(3) =
There are 8 − 2 + 1 = 7 terms, the first term is 2(3) = 18, and the common ratio is r = 3. Using the
formula we find that 18
= 18 = 19674. Geometric series start to get out of hand pretty
quickly, so the formula will come in handy for larger groups of terms.
Example 2
4 + 16 + 64 + 256 + ⋯ + 4194304 =
In this particular case, we know the first term t = 4 and the common ratio r = 4, but we do not know
the number of terms. Using skills gained in II.E.2, we find that
t = 4(4) = 4194304
4 = 1048576
(n − 1) log 4 = log 1048576
n =
+ 1 = 11 .
Therefore, the sum equals 4
= 5592404.
Note: Notice that the sum of a geometric series, in the previous case 5592404, is not significantly
different from the largest term in the series, 4194303 if you take into account that the series went from
4 to 4 million in only 11 terms. This conclusion will be most significant when we calculate an infinite
geometric series.
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Example 3
If ∑
256
=
, what is b?
The first term t = 256 = 128, the common ratio r = , and there are b terms. Therefore the sum
equals 128
=
=
1 − =
()
= 1 −
()
b =
()
= 15. HOLY FRIJOLES!
Infinite Geometric Series
An infinite geometric series may sound more difficult than what we’ve done up until now, but it’s
actually a ton easier. There is one very, very important stipulation: |r| < 1 to calculate an infinite
geometric series, otherwise the value gets to large, and we say it’s divergent. What may come as
shock to most people, is that the infinite sum of a geometric sequence is actually a finite value. Since
the ratio is so small, the terms get smaller and smaller, so the sum, itself, doesn’t change a whole lot
once we get far enough along. In the previous example, the first term was 128 and the sum was
= 255.9921875, almost twice the first term. In fact, if we had increased the number of terms in
this series, we would get closer and closer to 256 without going over it. The infinite geometric series
converges at 256 exactly. Since infinity is only theoretical, and we can never reach it, we can’t
actually find the infinite sum, but we can get close. If |r| < 1 and b → ∞, then (r) gets so small, it
eventually disappears. So the formula for a geometric series goes from
t (r)
= t 1 − r
1 − r
to t (r) = t 1 − r
1 − r = t 1
1 − r
= t
1 − r
Example 4
18 2
3
=
t = 18 = 8, r =
< 1, so the infinite sum is
=
= 8(3) = 24.
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II.E.4 Independent Practice
Calculate the following Geometric Series
1.
3(2) =
6.
4(5)
=
2.
2(3) =
7.
20 1
2
=
3.
5(2) =
8.
64 3
4
=
4.
8 1
2
=
9.
6 2
3
=
5.
5(3) =
10.
10 3
5
=
11. 2 + 4 + 8 + 16 + 32 + ⋯ + 256
14. 20 + 10 + 5 + + + ⋯
12. 1 + 3 + 9 + 27 + 81 + ⋯ + 1594323
15. 80 + 16 + +
+
+ ⋯
13. 2048 + 512 + 128 + 32 + 16 + ⋯ +
16. 100 + 75 +
+
+
+ ⋯
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Section
II.E.5
Applications of
Sequences & Series
We will use our new knowledge of sequences and series to answer the following word problems.
2.E.5 Independent Practice
Solve the following problems.
1. John is breeding rabbits. At the beginning of every month he has twice as many rabbits as he
had the month before. If he starts out with 10 rabbits and after the first month he has 20
rabbits. How many rabbits will he have after 12 months?
2. Dr. Brooks has a rare nuclear isotope with a half-life of 5 minutes. If he has 100 grams of the
isotope at noon, then he will have 50 grams at 12:05. How many grams will have left at 1:00?
3. Jenny’s parents bought her a cell phone and pay her bill each month. Her cell phone plan has
unlimited minutes for a base price plus an extra fee per text message. She sent 120 text
messages last month and received a bill for $43. This month she sent 235 text messages and
received a bill for $60.25. How much is the base price? How much is she charged per text
message? How much will she be charged for 900 text messages?
4. Bill tries to make a deal with his parents. He will take a penny for an allowance this week if his
parents will triple his allowance each week. How much would he get the fifteenth week? If he
does not spend any money, how much will he have total by the fifteenth week?
5. A vat contains 500 pounds of sugar. One minute later, 250 pounds are dumped in to the vat.
A minute after that 125 pounds are dumped in. If this pattern continues, how many pounds of
sugar would be dumped in 20 minutes after the process began. How many total pounds of
sugar will be contained within the vat after 20 minutes? If the process continues forever, how
many total pounds will be contained within the vat?
6. Bill makes another deal with his parents. He will take a dollar for an allowance this week if his
parents will add 25 cents to his allowance each week. How much would he get the fifteenth
week? If he does not spend any money, how much will he have total by the fifteenth week?
7. If Gertrude starts out with 30 million dollars, spends no money, and earns 15 percent interest
each year. How much interest will she earn 10 years later? How much total interest will she
earn in the first 10 years? If Gertrude lives forever, how much total interest will she earn?
8. Gordon buys two hot dogs. The next day he purchases 5 hot dogs. The day after that he
purchases 8 hot dogs. If this pattern continues, how many hot dogs will he purchase on the
thirtieth day? How many hot dogs will he have purchased total in thirty days?
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Unit
III
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Unit
III Unit Overview
The Purpose of the Unit
The purpose of this unit is to review and solidify the concepts that go into geometry, specifically
relating to triangles in preparation of Trigonometry. We will study right triangles at length, examining
SOH CAH TOA from a completely new point of view, and calculate the areas of non-right triangles.
We will then apply the knowledge of triangles to physics applications concerning two-dimensional
vectors.
This History of Geometry
The word ‘geometry’ is of Greek decent that means measure of the earth. Although geometry can
literally be used to calculate the measurements of the planet, itself, the name implies the natural
aspects of geometry and its uses to define mathematical relationships in every day phenomena.
This naturalistic aspect of geometry is best illustrated by its early use by cavemen in the ancient
Indus Valley. The ancient Babylonians and Egyptians both worked extensively with geometry through
surveying, construction, and astronomy. Both civilizations were aware of the Pythagorean Theorem
almost 1500 years before Pythagoras even walked the earth. The Babylonians and Egyptians,
working with circles, both developed a ratio between the diameter and circumference that would
eventually become pi. Even the Old Testament mentions the ratio being about 3.
Geometry really took off when the Greeks got their hands on it. Greece is credited with modern-day
geometry, so much so that we use a Greek word to name it and Greek letters to stand for geometric
variables. We even use a Greek letter, π, to stand for one of the most significant constant ever. One
of the first Greeks to delve into the study of geometry was Pythagoras and his followers the
Pythagoreans. The Pythagorean Theorem is named for him because he was the first to successfully
prove it. The Greek philosopher had a hand in geometry as well, stating all geometry should be done
with a ruler and compass.
No one had a bigger impact on Geometry than Euclid. Euclid determined most of the geometric
relationships that we understand today, and he wrote them into 13 books entitled The Elements of
Geometry. This book is still available today over 2300 years later. The geometry we learn in
elementary, middle, and high school is known as Euclidean geometry due to this mathematician.
Archimedes came next, and he is often considered one of the three greatest mathematicians of all
time due to the fact that he developed the beginnings of analytic or coordinate geometry and calculus
in the 2 nd century BCE, which wouldn’t be completely defined until almost 2000 years later.
In the 17 th century with the development of Algebra well under way, French mathematician, Rene
Descarte and Pierre de Fermat defined analytic geometry that sets geometric figures on a coordinate
grid. This branch of geometry would eventually lead to the development of calculus by one of the
other three greatest mathematicians, Isaac Newton, and his competitor, Gottfried Liebnitz later that
century. The last of the “Great Three”, Johann Carl Friedrich Gauss, would delve into non-Euclidean
geometry and branch out into other studies of mathematics.
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Chapter
III.A
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Chapter
III.A Chapter
Overview
‣ III.A.1 – The Pythagorean Theorem, Distance Formula,
& Equation of a Circle
This section investigates the close relationship between the Pythagorean Theorem, Distance
Formula, and the Equation of a Circle.
‣ III.A.2 – Sine, Cosine, & Tangent
This section begins with our pre-knowledge of SOH CAH TOA, and redefines it in a easy-tounderstand
method that has implications for the study of Trigonometry in later chapters.
‣ III.A.3 – Special Right Triangles
This section examines the unique relationships found in 45-45-90 and 30-60-90 triangles, using
their geometric construction to create easy-to-remember formulas.
‣ III.A.4 – Triangle Area Formulas
This section provides formulas for finding the areas of oblique triangles, as well as proof.
‣ III.A.5 – Triangle Applications
This section combines all knowledge gained throughout the chapter to answer application
questions.
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Section
III.A.1
The Pythagorean Theorem, Distance
Formula, & Equation of a Circle
This section has the longest title in the entire text, and yet, all three of the concepts are pretty much
the same thing. So shall we begin?
The Pythagorean Theorem
The Pythagorean Theorem is notable for two things. (1) It’s the most memorized mathematical
formula ever, and (2) it was horribly misquoted by the Scarecrow at the end of the Wizard of Oz.
“The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of
the remaining side,” is so ridiculously wrong that it’s proof positive that a diploma does not equate to a
brain. The correct definition is the sum of the squares of the legs of a right triangle is equal to the
square of the hypotenuse, or more commonly, a + b = c , given that a and b are the legs and c is
the hypotenuse of a right triangle.
Example 1
Calculate x.
x
7
3
Since x is the length of the hypotenuse, 3 + 7 = x , and x = √3 + 7 = √58. Even though, solving
for x should yield a negative value,algebraically, there is no such thing as negative length in
geometry.
Example 2
Calculate x.
7
x
3
Since x is now the length of a leg, x + 3 = 7 , and x = √7 − 3 = √40 = 2√10.
The Distance Formula
The distance formula is one of the basic ideas of analytic geometry. The title, however, of this
segment is actually a misnomer, as we are not going to use a formula, mainly because no one ever
remembers the distance formula. Given two points (x , y ) and (x , y ), we will find the horizontal
distance and vertical distance between the two points, assign those lengths to the legs of a right
triangle, and calculate the hypotenuse.
Example 3
Find the distance between the points (2,5) and (-4,8).
The horizontal distance between the points is 6 and the vertical distance between the points is 3.
Now, all we have to do is calculate the hypotenuse, d = √6 + 3 = √45 = 3√5.
Note: It does not matter where the points are in terms of one another. If we think of this in terms of a
right triangle and the Pythagorean Theorem, it will be easier when we discuss vectors in III.B.
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Equations of Circles
How do we define a circle? Do we define it as a perfectly round geometric figure? That’s how almost
everyone in the world describes it, but if we truly understand the geometric make-up of a circle,
creating an equation for one becomes easier. A circle is the set of all points that are equidistant
from a single point. The single point is the center, and the distance every point is from the center is
the radius. A few quick distinctions: the circle is made up of an infinite number of points, and the
center is not part of the circle; it’s the reference point.
Example 4
Determine the equation of a circle centered at the origin with a radius of 5.
Every point (x,y) on the circle is a distance of 5 from the origin (0,0) as illustrated by
5
the diagram. Notice we can make a right triangle out of this information. The
x
(0,0)
horizontal distance is x − 0 = x, and the vertical distance is y − 0 = y. According to
the Pythagorean Theorem, x + y = 25. Coincidentally, this is the equation of the
circle, since every point that satisfies the equation is of a distance 5 from (0,0), the definition of a
circle.
y
(x,y)
Example 5
Determine the equation of a circle centered at (2,-5) with a radius of 7.
Every point (x,y) on the circle is a distance of 7 from (2,-5). The horizontal distance
would be x − 2 and the vertical distance is y + 5. Applying the Pythagorean
Theorem, (x − 2) + (y + 5) = 49.
7
x-2
(2,-5)
(x,y)
y+5
As we can see, the Pythagorean Theorem has multiple uses, and we’ve only touched the surface.
This simple theorem will reappear; it’s guaranteed.
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III.A.1 Independent Practice
Solve for x.
x
3
x
10
1.
4
3.
12
x
13
x
8 2
2.
5
4.
16
Calculate the distance between the following points.
5. (3,4) & (2,7)
8. (3,9) & (0,-2)
6. (5,6) & (-3,4)
9. (-3,-5) & (2,3)
7. (-2,4) & (4,-1)
10. (8,6) & (4,-7)
Given the center and radius for a circle, determine its equation.
11. Center: (0,0) & r = 5
14. Center: (4,-2) & r = 3
12. Center: (3,7) & r = 2
15. Center: (-5,-9) & r = 12
13. Center: (-2,6) & r = 7
16. Center: (2,0) & r = 9
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Section
III.A.2
Sine, Cosine,
& Tangent
Back in geometry, we learned about sine, cosine, and tangent, and we did so with the help of the
phrase SOH CAH TOA. The one disadvantage to this is we have to remember the correct spelling
and remember what it actually means. To refresh:
Sine =
O
H , Cosine =
A
H
, Tangent =
O
A
As with much of this text, we’re going to throw this method away after redefining sine, cosine, and
tangent. But first… let’s go back to the basics.
Basics
Angles
o A triangle’s angles add up to 180°.
o A right triangle has one 90° angle.
o So, the two acute angles in a right triangle add up to 90°.
Sine, Cosine, and Tangent
o sin, cos, and tan are functions.
o The input, θ, is an angle.
o The outputs, sin(θ) , cos(θ) , and tan(θ) are ratios.
Inverse Sine, Inverse Cosine, and Inverse Tangent
o sin , cos , and tan are functions.
o The inputs, sin θ , cos θ, and tan θ are ratios.
o The output, θ, is an angle.
Old School Method
If we have a right triangle with an acute angle of focus θ as shown, the ratios
are sin(θ) =
, cos(θ) = , and tan(θ) = .
hyp
adj
opp
Example 1
Calculate the sine, cosine, and tangent of θ.
5
4
3
sin θ = , cos θ = , and tan θ = .
Let’s get something straight: there’s nothing wrong with the old school method, but it’s very difficult to
link this method to future chapters, which is why we’re going to change it up a bit.
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New Method
Before we do any right triangle problems, we have to set the problem up correctly,
1
and if we set every triangle in the same manner, we don’t have to remember any
silly acronyms. We need the triangle set up like the diagram, such that the acute
angle of focus is in the bottom left position and the hypotenuse equals 1.
1
Once the triangle is ready, the cosine is the length of the horizontal leg, the sine is the
length of the vertical leg, and the tangent is the slope (not the length!) of the
hypotenuse (tan θ =
cos()
). Also, we will refer to these as trig ratios and trig functions.
Example 2
13
12
Calculate the sine, cosine, and tangent of θ.
5
1
5/13
Firstly, we have to set up the triangle correctly, which involves flipping the triangle
12/13
and dividing all three sides by 13 so the hypotenuse will equal 1. Based on the
triangle: cos θ =
, sin θ =
, and tan θ =
= , since the 13’s cancel out.
Note: An easy way to remember that cosine is horizontal and sine is vertical is that (cos,sin) is
alphabetical just like (x,y). The hypotenuse equaling 1 is also significant, but we won’t reveal the
significance here. We’ll see it in Unit IV.
Example 3
X
13
y
Solve the triangle:
25
x
X
The triangle is already in standard position, but the hypotenuse doesn’t equal 1.
1
y/13
We need to divide everything by 13, giving us the triangle to the right. Firstly,
X = 90° − 25° = 65°. Next, cos 25° = 25
x/13
→ x = 13 cos 25° =11.782,
sin 25° = → y = 13 sin 25° = 5.494.
Note: To get the correct values in the calculator, we must put the calculator in DEGREE in the MODE
menu. The default (every time we reset the calculator) is RAD, which we’ll talk about in Unit IV.
Example 4
h 52
6
Solve the triangle:
Y
x
1 52
Y = 90° − 52° = 38°.
6/h
sin 38° = → h =
= 9.746
38
° x/h
tan 38° = → x =
= 7.680. °
sin()
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Examples 3 and 4 demonstrate several interesting facts. In example 3, we didn’t use tangent
because tan 25° = , and we can’t solve for 2 variables at once. In example 4, we didn’t use cosine
for the same reason. Additionally, in Example 3, the ratios were a variable over a constant, so the
variable was equal to the constant times the trig ratio. However, in Example 4, the ratios were a
constant over a variable, so the variable was equal to the constant divided by the trig ratio. We can
always solve these with cross multiplication.
Inverse Trig Functions
Remember, inverse functions take in the original output and create the original input. Since trig
functions take in an angle and produce a ratio, the inverse trig functions will take in a ratio and
produce an angle. So, if we need to find an angle in a right triangle, we know we have to use an
inverse trig function.
Example 5
Solve the triangle:
Y
h
5
X
4
Firstly, we should be able to calculate the hypotenuse using the Pythagorean Theorem.
h = √4 + 5 = √41. Based on the values we’ve been given, tan Y = . Therefore,
Y = tan = 38.660°, and X = 90° − Y = 51.340°. Notice that we did not redraw the triangle.
Since the only trig ratio we were using was tangent, and the slope doesn’t depend on the length of
the hypotenuse, there’s no reason to make the hypotenuse equal to 1.
Example 6
Solve the triangle:
x = √8 − 3 = √55
X
8
3
Y
x
sin Y = 3 8 → Y = sin 3 8 = 22.024°
X = 90° − Y = 67.976°
Y
1
x/8
X
3/8
Note: We make sure to write the degree symbol next to every angle measure to distinguish it from
other angle measuring methods, which we will talk about in Unit IV. We need to get in the habit of
doing this now.
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III.A.2 Independent Practice
Solve for a, b, and c in the following triangles.
5
a
c
22
29
c
1.
35
b
6.
b
a
b
a
c
c
72
11
2.
35
3
7.
b
a
b
a
4
c
67
b
3.
58
c
8.
a
6
8
a
4
b
c
9
4.
b
c
9.
a
4
13
a
c
15
c
9
5.
b
8
10.
a
b
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Section
III.A.3 Special
Right Triangles
There are two types of triangles that we consider “special”, not to discourage other triangles or injure
their precious and fragile self-esteem. These two types are called 45-45-90 and 30-60-90 triangles
for fairly straightforward reasons. The 45-45-90 triangles have interior angles of 45°, 45°, and 90°,
whereas the 30-60-90 triangles have interior angles of 30°, 60°, and 90°. See? Straightforward…
45-45-90 Triangles
The thing that makes these special is the fact that they are halves of squares. In
bisecting the square in this way, we bisect the right angles creating (90°) = 45° angles.
The unique qualities of the 45-45-90 triangle is that both legs are equal in length (x), and
the hypotenuse is of length √x + x = √2x = x√2, by the Pythagorean Theorem.
Example 1
Calculate the length of the hypotenuse of a 45-45-90 triangle if the leg length is 3.
If x = 3, then the hypotenuse equals x√2 = 3√2.
Example 2
Calculate the length of a leg of a 45-45-90 triangle if the hypotenuse is of length 8.
If x√2 = 8, then x =
√ = 4√2.
45
x 2
x
45
x
Now, if we set our 45-45-90 triangle in standard form as the diagram to the right shows,
we find that the hypotenuse x√2 = 1. Therefore the leg x =
. Since the cosine
= √
√
is the horizontal length and sine is the vertical length, cos 45° = √
tan 45° =
°
° = 1.
, sin 45° =
√
, and
45
1
45
Example 3
Calculate the length of the hypotenuse of a 45-45-90 triangle if the length of a leg is 7√3.
Since we know the length of the leg is either cosine or sine, then cos 45° = √
= √
.
Through cross-multiplication, we get h√2 = 14√3 → h = √
√
= √
= 2√6. This method
is not completely necessary, but if we understand it, we’re better positioned to perform better in trig.
45
1
7 3
h
45
7 3
h
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30-60-90 Triangles
The special properties of a 30-60-90 triangle derives from the fact that it is half of an
equilateral triangle (all sides are equal and all angles are 60°). Bisecting the triangle,
we use a segment that is both an altitude (creates a right angle) and angle bisector
(creates a 30° angle). For this reason the short leg of a 30-60-90 triangle (x) is one half
of the hypotenuse (2x), and the long leg (opposite the 60° angle) is (2x) − x =
√4x − x = √3x = x√3.
30
2x
x 3
60
x
Example 4
Given a 30-60-90 triangle with a short leg of length 8, what are the measurements of the other two
lengths?
Since x = 8, the long leg x√3 = 8√3, and the hypotenuse 2x = 16.
Example 5
Given a 30-60-90 triangle with a hypotenuse of length 7, what are the measurements of the other two
lengths?
Since, 2x = 7, then the short leg x = , and the long leg x√3 = √
√3 = .
Setting the 30-60-90 triangle in standard position in terms of the 30° angle as it is at
right, the hypotenuse 2x = 1. Therefore, the short leg (sin 30°) x = , and the long leg
(cos 30°) x√3 = √ . Likewise if we set the same triangle in standard position in terms
of 60° with hypotenuse 2x = 1, then the short leg (cos 60°) x = , and the long leg (sin 60°)
30
1
1
60
30
x√3 = √
.
60
Example 6
If the long leg of a 30-60-90 triangle is 5√2, what is the length of the short leg?
Let’s set the triangle in standard position in terms of 30°. Since we are solving for x,
there is no need to use h. To eliminate h, we can use tangent. Then tan 30° =
√
. We cross multiply, and x√3 = 5√2 → x = . Like example
°
=
°
= √ √
= √
√
3, if we understand the method employed in this example, we are going to be better
prepared to master trig.
30
1
5 2
h
60
x
h
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III.A.3 Independent Practice
Solve for a and b.
a
4
7 2
a
1.
45
b
4.
45
b
a
5 2
3 5
a
2.
45
b
5.
45
b
a
5 3
6
a
3.
45
b
6.
45
b
a
60
5
6
60
b
7.
30
b
11.
30
a
a
60
4 3
10 3
60
b
8.
30
b
12.
30
a
a
60
b
1
60
b
9.
30
8 3
13.
30
a
a
60
b
1
a
10.
30
9
14.
45
b
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Section
III.A.4
Triangle Area
Formulas
Hopefully, by now, we are familiar with the formula for finding the area of a triangle A = bh, where b
is the base length, and h is the height (length of the altitude perpendicular to the base) as shown in
the following three triangles.
h
h
b
b
h
b
Example 1
Given the triangle, find its area.
Since b = 20 and h = 3, A = bh = (20)(3) = 30.
3
20
The only problem with this formula is that we must know both the length of one of the side lengths
and the length of the altitude to that side length from the opposite vertex. That seems like a little too
much to rely upon. There are several area formulas of which we can make use depending on what
parts we are given.
Triangle Congruency
In certain cases, there is only one possible triangle that has a particular set of measurements in a
particular order. Therefore, any two triangles with those measurements in common must actually be
congruent in both size and shape. Two triangles are congruent:
If they share the same three side lengths. (SSS)
If they share two side lengths and the angle formed by those sides. (SAS)
If they share two angles and a corresponding side length. (AAS or ASA)
If they are right triangles that share the same hypotenuse and leg. (HL)
Two triangles are not necessarily congruent:
If they share two angles with unknown side lengths. (AA can prove similarity, same shape /
different size)
If they share two sides with an angle not between them. (SSA… Don’t spell this backwards)
We can only find the area of a triangle if it fits one of these cases. If we can’t prove that only one
possible triangle exists with these measurements, then we can’t guarantee a correct value of the
area.
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SSS Triangles
If we know all three sides of an oblique triangle, then we can use Hero’s (or Heron’s) Formula:
Area = s(s − a)(s − b)(s − c)
B
a
C
b
Where s =
stands for the semi-perimeter. This will be proven in IV.C.2.
c
A
Note: Not all sets of three lengths can be the side lengths of a triangle. If a ≤ b ≤ c, then a + b > c in
a triangle.
Example 2
Given a triangle with side lengths 5, 8, and 9, what is the area of the triangle?
Since, 5 + 8 > 9, then s =
= 11, and so,
Area = 11(11 − 5)(11 − 8)(11 − 9) = 11(5)(3)(2) = √330.
SAS Triangles
If we know two sides and the angle between them of an oblique triangle, then we can use the
formula:
Area = 1 ab sin C
2
We can prove this by dropping the altitude (unknown length h) to side length b in the
triangle to the right. We can find the length of h if we look at just the right triangle
created by side lengths a, h, and part of b. Dividing all sides by a, making the
hypotenuse 1, we find that sin C = and, thus, h = a sin C. Since Area = bh, we get
C
a
b
h
Area = b(a sin C) = ab sin C. Q.E.D.
Example 3
In triangle PQR, p = 3, q = 7, and R = 37°, what is the area of the triangle?
Area = pq sin R = (3)(7) sin 37° = 6.319.
This formula is actually the basis for the formula use in the next case, but the proof will be left until
VI.C.
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AAS and ASA Triangles
If we know any side length and two angles, we essentially know one side length and
all three angles. We can find the third angle by subtracting the two we know from
180°. If we know side a and all three angles A, B, & C, then we can use the formula:
B
a
C
b
Area = a sin B sin C
2 sin A
c
A
This is possible because b =
by the Law of Sines we will investigate in IV.C.1.
Example 4
In triangle KLM, K = 82°, L = 41°, and k = 6. What is its area?
Firstly, M = 180° − K − L = 180° − 82° − 41° = 57°.
So, Area =
= ° °
= 10.001.
°
Note: The angle opposite the known side always goes in the denominator.
Calculator Note: Remember to put the entire numerator and denominator in their own sets of
parentheses.
Right Triangles
In order to calculate the area of a right triangle, we must know the lengths of both legs. Once we do,
we make one leg the base and the other, the height. Then we use Area = bh… pretty simple…
Example 5
Find the area of a right triangle with a hypotenuse of 8 and a leg of 5.
Using the Pythagorean Theorem, the other leg is √8 − 5 = √39. Therefore,
Area = (5)√39 = 15.612.
Example 6
Find the area of a right triangle with acute angle 48° and hypotenuse of length 4.
We need to find the two legs l and l . l = 4 cos 48° and l = 4 sin 48°. Therefore,
Area = (4 cos 48°)(4 sin 48°) = 3.978.
Note: We could have used the AAAS formula as well ° °
°
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= 3.978.
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III.A.4 Independent Practice
Calculate the areas of the triangles.
7 5
1.
45
5
7.
9
3 9
20
8.
10
2.
45
4
3.
30
8
9.
23
5
7
4.
30
12
10.
52
10
10
5.
55
11.
83
9
24
42
8
6.
6
12.
31
123
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Section
III.A.5 Triangle
Applications
In this section, we will solve word problems concerning triangles, their angles, and their areas.
III.A.5 Independent Practice
Draw a picture and solve the following questions.
1. Jordan is standing on level ground. The sun is currently at a 50° elevation. If Jordan is 6 feet
tall, how long is his shadow?
2. Later in the day, 6-foot-tall Jordan’s shadow is 12 feet long. What is the sun’s elevation?
3. If the base of a ramp is 16 feet from a door, which is 2 feet off the ground, how long is the
actual ramp? What is the elevation for the ramp from ground level?
4. If Charles walks 5 miles east and 15 miles north, how far is Charles from his starting position?
5. If 6.5 foot tall Wanda is standing 20 feet from a tree and looking with an angle of elevation of
36° at the top of the tree, how tall is the tree?
6. Hank is standing at the top of a 500 ft tall building and is looking down at an angle of
depression of 42° at a 350 ft tall building. How far apart are the two buildings?
7. Xander wants to build a triangular enclosure for his pigs. If the side lengths are 20 yd, 30 yd,
and 40 yd, what is the overall area of the enclosure?
8. Paula is a surveyor. She stands at one corner of an oblique triangular lot and determines the
angle she’s standing at is 123°. If the other two corners are 60 ft and 42 ft from her corner,
what is the overall area of the lot?
9. Jerry walks east 5 miles, north 7 miles, west 2 miles, north 9 miles, east 6 miles, and south 1
mile. How far is the now-lost Jerry from his starting position?
10. Tammy starts her car at sea level. If Tammy drives 5 miles up an incline of 30 degrees, how
many miles above sea level will she be at the end of the drive?
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Chapter
III.B
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Chapter
III.B Chapter
Overview
‣ III.B.1– Component Form, Magnitude, & Direction
This section defines a vector in terms of its magnitude and direction, and then rewrites the vector
in terms of its horizontal and vertical components. It also applies III.A to coordinate (analytic)
geometry.
‣ III.B.2 – Vector Addition
This section supplies the means and knowledge to add multiple vectors together.
‣ III.B.3 – Dot Product
This section explores the concept of dot product and calculates the angle between two vectors.
‣ III.B.4 – Vector Projection
This section determines, given two vectors, the magnitude applied by one vector in the direction of
the other, thereby creating a third vector.
‣ III.B.5 – Physics Applications
This section combines the previous four sections, allowing for the solving of real-world physicsbased
mechanics questions.
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Section
III.B.1
Component Form,
Magnitude, & Direction
What is a vector?
A vector is a segment that has both length and direction. The direction of the vector is implied by an
arrow, which we refer to as the head of the vector, and we call the other end the tail. The length of a
vector is known as the magnitude, and it can represent displacement, velocity, acceleration, force,
and any other measurement that contains direction. Distance and speed cannot be represented as
vectors because they contain only magnitude and are always considered positive. The direction is
any degree measure between 0° (the positive x-axis) and 360° (also the positive x-axis after one
counterclockwise revolution).
180
t
90
w
v
0
270
360
This diagram to the left shows four vectors, u⃗, v⃗, w⃗, and t⃗. v⃗ is a position vector
whose tail is located at the origin with a magnitude, |v⃗| = 3 and a direction
θ ⃗ = 45°. w⃗ is a displacement vector with a magnitude, |w⃗| = 3 and a direction
θ ⃗ = 45°. Since v⃗ and w⃗ share the same magnitude and direction, v⃗ = w⃗,
regardless of their location. t⃗ also has a magnitude, t⃗ = 3, but its direction is
θ ⃗
= 225°, an exactly 180° difference from v⃗. Since the magnitudes are equal,
but the directions are exactly opposite, we say t⃗ = −v⃗. u⃗ is a unit vector with
magnitude, |u⃗| = 1 and direction, θ ⃗ = 150°.
Component Form
Every vector can be broken down into its horizontal and vertical components using the trigonometric
functions learned in the last chapter. Unit vector, ı⃗, has a magnitude, |ı⃗| = 1, and a direction, θ ⃗ = 0°
(positive x-axis), and unit vector, ȷ⃗, has a magnitude |ȷ⃗| = 1, and a direction, θ ⃗ = 90° (positive y-axis).
Therefore, a vector written as 3ı⃗ − 4ȷ⃗ has a horizontal displacement of right 3 and vertical
displacement of down 4.
Example 1
Calculate the magnitude and direction of v⃗ = 3ı⃗ − 4ȷ⃗.
90
Based solely on the graph, we see that v⃗, 3ı⃗, and −4ȷ⃗ make up a right triangle.
Therefore, we can use the Pythagorean Theorem: |v| = √3 + 4 = 5. The
direction is a little trickier. Remember, the tangent of an angle is equal to the
slope of the hypotenuse. Therefore tan θ ⃗ = − , and thus θ ⃗ = tan − =
270
−53.130°. A negative angle indicates clockwise rotation. Therefore this angle
represents a vector in Quadrant IV, so it’s correct. The problem, however, with this angle is that it
needs to be between 0° and 360°, so we need to add one more revolution to it by adding −53.130° +
360° = 306.870°.
3ı⃗ − 4ȷ⃗ can also be written with angled brackets: 〈3,4〉. This is typically easier and quicker notation to
write and grasp. Additionally, if 〈3,4〉 is a position vector with its tail at (0,0), then its head is at (3,4).
180
3i
v
-4j
0
360
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Example 2
Calculate the magnitude and direction of t⃗ = 〈−2,5〉.
We can create a position vector as we did in Example 1 whose head is at (-2,5) in the second
quadrant. Therefore t⃗ = (−2) + 5 = √29. We know because the vector is in Quadrant II, its
direction must be between 90° and 180°. Since tan θ ⃗
=
, then θ ⃗ = tan
= −68.199°. This
measurement points to a vector in Quadrant IV, which is wrong. So we add 180°. Therefore,
θ ⃗ = −68.199° + 180° = 111.801°.
Note: Two vectors 180° apart have the same slope and, thus, the same tangent. If the inverse
tangent results in a negative angle in the correct quadrant, we add 360°, but if it results in an angle in
the wrong quadrant, altogether, we add 180°.
If we are given the magnitude and direction of a vector and want to determine the vector’s component
form, we need only take a peek back at III.A.
Example 3
Determine the component form of a vector with a magnitude of 3 and a direction of 240°.
After drawing a diagram of a position vector in the third quadrant and its
components, we see that we have a right triangle with a known hypotenuse. If we
divide all three sides by 3, the hypotenuse converts to 1, and the sides convert to
cos 240° = ⃗
⃗
and sin 240° = . Therefore xı⃗ = 3 cos 240° = −1.5 and
yȷ⃗ = 3 sin 240° = −2.598. Therefore the vector in component form is
〈−1.5, −2.598〉.
180
yj
240
xi
3
90
0
270
360
v⃗ = 〈|v⃗| cos θ ⃗ , |v⃗| sin θ ⃗ 〉
Displacement Vectors
A displacement vector is the vector that points from one point to another point. To find it we subtract
the head point minus the tail point.
Example 4
ab ⃗ = 〈x − x , y − y 〉
Find the magnitude and direction of the displacement vector that points from a(1,5) to b(−2,3).
The position vector ab ⃗ = 〈−2 − 1,3 − 5〉 = 〈−3, −2〉. Therefore, (ab) ⃗ = (−3) + (−2) = √13, and
the direction equals tan
= 33.690° → 33.690° + 180° = 213.690°.
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III.B.1 Independent Practice
Determine displacement vectors, ab ⃑, their magnitude, and direction.
1. a(3,6) & b(5,8)
5. a(2, −5) & b(4, −3)
2. a(−4,6) & b(2,5)
6. a(−1,7) & b(5,2)
3. a(5,3) & b(2,0)
7. a(4,2) & b(−2,5)
4. a(9,4) & b(13,5)
8. a(5,4) & b(2,1)
Determine vectors in component form given their directions and magnitudes.
9. |a⃗| = 5; θ = 70°
13. |e⃗| = 6; θ = 347°
10. b⃗ = 3; θ = 123°
14. f⃗ = 12; θ = 250°
11. |c⃗| = 4; θ = 36°
15. |g⃗| = 2; θ = 140°
12. d⃗ = 8; θ = 195°
16. h⃗ = 9; θ = 355°
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y
(2,5 )
(7,-5)
x
Section
III.B.2 Vector
Addition
Vector addition is extremely simple as long as we adhere to the following guidelines:
Added vectors must be placed head-to-tail.
The sum of the vectors, or resultant, connects the first tail to the last head.
If the vectors are in component form, the vectors add like matrices.
a
a
a+b
b
b
a+b
b
a
The first diagram shows a⃗ and b⃗. The second diagram connects the head of a⃗ to the tail of b⃗, and the
third connects the head of b⃗ to the tail of a⃗, both of which result in the same a⃗ + b⃗, indicating that
vector addition is commutative.
Example 1
If a⃗ has a magnitude of 3 and a direction of 60°, and b⃗ has a magnitude of 6 and a direction of 135°,
calculate a⃗ + b⃗, a⃗ + b⃗, and θ ⃗ ⃗.
a⃗ = 〈3 cos 60° , 3 sin 60°〉, b = 〈6 cos 135° , 6 sin 135°〉, and
a⃗ + b⃗ = 〈3 cos 60° + 6 cos 135° , 2 sin 60° + 6 sin 135°〉 = 〈−2.743,5.974〉.
a⃗ + b⃗ = (−2.743) + 5.974 = 6.574. θ ⃗ = tan .
+ 180° = 114.657°.
.
Note: It may benefit to utilize the STO> key when calculating the components of a⃗ + b⃗. Then we can
use the exact values in the calculator to solve the rest of the problem, rounding to three decimal
places at the end.
Example 2
Determine the point that is 60% of the way from (2,5) to (7,-5).
Essentially we are looking for a position vector that points to a point 60% of the
way from (2,5) to (7,-5). To do that we will add the position vector 〈2,5〉 to 60% of
the displacement vector 〈7 − 2, −5 − 5〉 = 〈5, −10〉. Thus 〈2,5〉 + .6〈5, −10〉 =
〈2,5〉 + 〈3, −6〉 = 〈5, −1〉. Therefore the point is (5,-1).
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III.B.2 Independent Practice
Given a⃗ = 〈3, 5〉 and b⃗ = 〈2, −7〉, calculate the following.
1. a⃗ + b⃗ =
5. −5b⃗ =
2. a⃗ − b⃗ =
6. 5a⃗ + 7b⃗ =
3. b⃗ − a⃗ =
7. 3a⃗ − 8b⃗ =
4. 3a⃗ =
8. 2b⃗ − 4a⃗ =
Calculate a⃗ + b⃗, a⃗ + b⃗, and θ ab .
9. a⃗ = 〈4,2〉 and b⃗ = 〈2,5〉.
11. |a⃗| = 5, θ = 68°, b⃗ = 7, and θ = 124°.
10. a⃗ = 〈−3,5〉 and b⃗ = 〈4, −2〉.
12. |a⃗| = 3, θ = 232°, b⃗ = 6, and θ = 23°.
Calculate the following.
13. The point 50% of the way from (3,2) and (5,7).
14. The point 25% of the way from (5, −2) and (2,8).
15. The point 60% of the way from (9,3) and (5,0).
16. The point 18% of the way from (−3,5) and (4, −1).
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Section
III.B.3 Dot Product
The dot product, or inner product, is the product of two vectors that results in a single numerical
value. This dot product has several essential applications, one of which we will discuss in this section
and one in the next.
Calculating the Dot Product
There are two ways to calculate the dot product of a⃗ = 〈a , a 〉 and b⃗ = 〈b , b 〉, but
to do so, we must position the vectors tail-to-tail with an angle between them θ.
a⃗
θ
b⃗
a⃗ ∙ b⃗ = |a⃗|b⃗ cos θ & a⃗ ∙ b⃗ = a b + a b
Example 1
Calculate the dot product 〈3,5〉 ∙ 〈2, −4〉.
3(2) + 5(−4) = 6 − 20 = −14.
Example 2
Calculate the angle, θ, created when a⃗ = 〈3,5〉 and b⃗ = 〈2, −6〉 are placed tail-to-tail.
a⃗ ∙ b⃗ = 3(2) + 5(−6) = −24
|a⃗| = 3 + 5 = √34
b⃗ = 2 + (−6) = √40
a⃗ ∙ b⃗ = |a⃗|b⃗ cos θ → θ = cos ⃗∙ = |⃗|⃗ cos −
= 130.601°
()()
Note: We don’t have to worry about adding 360° or 180° in these problems like we did in III.B.1
because we’re not calculating direction. We’re calculating the angle between two vectors, which has
a maximum possible value of 180°. The angle we get in the calculator will be the correct answer,
guaranteed.
Unit Vectors
As established in the previous section, a unit vector is a vector that has a magnitude of 1 and simply
points in a direction.
Example 3
Calculate the unit vector in the direction of 〈5, −6〉.
|〈5, −6〉| = 5 + (−6) = √61. So the unit vector is 〈,〉
√
= 〈√ , − √ 〉.
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III.B.3 Independent Practice
Calculate the unit vectors of the following.
1. 〈3,5〉
4. 〈−2,6〉
2. 〈5, −12〉
5. 〈8,4〉
3. 〈7,4〉
6. 〈−4, −9〉
Calculate the dot products of the following.
7. 〈3,5〉 ∙ 〈2,4〉 =
10. 〈5,7〉 ∙ 〈6, −3〉 =
8. 〈1,7〉 ∙ 〈−3, −6〉 =
11. 〈21,4〉 ∙ 〈−3,5〉 =
9. 〈−2,6〉 ∙ 〈−5,4〉 =
12. 〈7, −2〉 ∙ 〈5, −1〉 =
Calculate θ, the angle between the two vectors when placed tail-to-tail.
13. 〈3,4〉 and 〈5,8〉
16. 〈−3,5〉 and 〈−5, −6〉
14. 〈3, −2〉 and 〈5,2〉
17. 〈5,6〉 and 〈−5, −6〉
15. 〈−5,1〉 and 〈3, −6〉
18. 〈4, −8〉 and 〈−5, −2〉
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Section
III.B.4 Vector
Projection
We already know that a vector can be broken into horizontal and vertical components, but vectors
have components in all sorts of directions, and that’s where vector projection comes into play.
As we can see, p⃗ is the vector projection of a⃗ in the direction of b⃗. There are
two steps to this: (1) find the magnitude, p = |p⃗| (the scalar projection), and
(2) the direction of b⃗, represented by the unit vector ⃗
⃗ .
a⃗
p = |a⃗| cos θ & p⃗ = p ∙ ⃗
The definition of p looks very similar to the dot product, a⃗ ∙ b⃗ = |a⃗|b⃗ cos θ, with
the |b⃗| divided out. Thus
Example 1
p = ⃗∙ ⃗
⃗
⃗
& p⃗ = ⃗∙ ⃗
⃗ b⃗
Calculate the scalar projection, p, of 〈5, −2〉 in the direction of 〈3,1〉.
a⃗ ∙ b⃗ = 5(3) − 2(1) = 13
b⃗ = 3 + 1 = √10
⃗∙⃗
⃗ =
√ = 4.111
Remember, this is the magnitude of the vector projection in the direction of 〈3,1〉.
Example 2
Calculate the vector projection, p⃗, of 〈−4,5〉 onto 〈2,1〉.
a⃗ ∙ b⃗ = (−4)(2) + 5(1) = −3
b⃗ = 2 + 1 = √5
p⃗ = a⃗ ∙ b ⃗
b⃗ b ⃗ = − 3
√5 〈2,1〉 = − 3 5 〈2,1〉 = 〈− 6 5 , − 3 5 〉
θ
p⃗
b⃗
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III.B.4 Independent Practice
Determine the scalar projection of a⃗ in the direction of b⃗.
1. a⃗ =< 3,4 > b⃗ =< 5,12 >
5. a⃗ =< 9,2 > b⃗ =< 5,4 >
2. a⃗ =< 3, −6 > b⃗ =< −6,2 >
6. a⃗ =< −2,6 > b⃗ =< −1, −9 >
3. a⃗ =< −2,7 > b⃗ =< 6,5 >
7. a⃗ =< 4, −3 > b⃗ =< 2, −8 >
4. a⃗ =< −2, −5 > b⃗ =< 3,7 >
8. a⃗ =< 8,5 > b⃗ =< 6,7 >
Determine the vector projection of a⃗ onto b⃗.
9. a⃗ =< 5,12 > b⃗ =< −3,4 >
13. a⃗ =< 6,1 > b⃗ =< −3, −9 >
10. a⃗ =< 8,2 > b⃗ =< 6, −2 >
14. a⃗ =< 7,2 > b⃗ =< −2,8 >
11. a⃗ =< 7,2 > b⃗ =< 8,5 >
15. a⃗ =< −2,5 > b⃗ < −3,8 >
12. a⃗ =< −2, −1 > b⃗ =< −3,5 >
16. a⃗ =< 5,3 > b⃗ =< −3, −7 >
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Section
III.B.5 Physics
Applications
We can now apply our new-found knowledge concerning vectors to solving physics word problems.
III.B.5 Independent Practice
Solve the following questions.
1. The autopilot on a plane is set to fly at 500 mph traveling at 30° north of east. A cross wind is
hitting the plane at 80 mph 65° north of east. What is the resultant speed and direction of the
plane?
2. The autopilot of another plane is set to fly at 650 mph traveling 35° north of west. A head wind
is hitting the plane at 65 mph 20° south of east. What is the resultant speed and direction of
the plane?
3. Ted walks 6 miles due north and 8 miles due east. The next day he walks 4 miles due south
and 2 miles due west. What is his resultant distance from his starting point? What direction
must he walk in to get back to his starting point?
4. A small heavy block needs to be moved but is too heavy to lift. You tie a rope to the top corner
and apply 20 newtons of force along the rope at an angle of 55° with ground level. How much
force is being applied parallel to the ground?
5. A ship is traveling at 20 knots 150° clockwise from north. The water’s current is moving at 5
knots 75° counterclockwise from north. What is the ship’s result velocity and direction?
6. If Sandra is driving up a 36° incline at 60 mph, how fast is she moving parallel to level ground?
7. Frank travels 6 miles west and 8 miles north from camp. Steve travels 2 miles east and 20
miles north from camp. If Sarah wants to travel to a point exactly ¾ of the way from Frank to
Steve, how does she get there?
8. A sign is hanging from two angled cables. If both cables are at an angle of 40 degrees with
the horizontal and each cable has a five pound tension on it, how much does the sign weigh?
9. George wants to cross a river that flows due south at 10 mph. In still water he can row at 15
mph. At what bearing must he row to end up due east of his starting point?
10. If a plane is traveling at 500 mph at 62° north of east with a tail wind of 72 mph at 70° north of
east, what wind velocity is actually being applied in the intended direction of the plane?
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Unit
IV
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Unit
IV Unit Overview
The Purpose of the Unit
The purpose of this unit is to introduce the basics of the branch of mathematics known as
trigonometry, the measurement of triangles. Yes, Unit III dealt with triangular geometry, but we are
now going to apply them to a coordinate grid (like vectors) and examine algebraic relevance of
trigonometric functions. This unit will help to prepare for any precalculus or calculus classes we may
take in college.
This History of Trigonometry
Trigonometry is Greek for “triangle measure”, and once again, we see the ancient Greeks’ knack for
stealing the thunder from other cultures. The study of triangles can be traced back to Egypt and
Babylonia more than 4000 years ago. Egypt used basic triangular knowledge to help build the
pyramids, assuming, of course, we don’t believe in alien-influenced architecture. The Babylonians
developed triangular mathematics to help define astronomy, possibly looking for alien life that could
come build pyramids for them. Hipparchus of Nicaea, a Greek mathematician, was labeled the
“Father of Trigonometry” for developing trigonometric tables, but was he really, or did he just have an
excellent publicist? Sure, Greece did devise the concept of the angle, ratios between side lengths,
and triangles in three dimensional space, but that’s where other cultures took the reins and took off.
Indian and Islamic astronomers and mathematicians made great strides towards what we know today
as modern Trigonometry. In fact India first devised the ideas of sine and cosine, which are Greek
translations of the Indian words, jya and kojya, respectively. Islamic mathematicians developed the
other trig ratios, tangent, secant, cosecant, and cotangent, ratios we haven’t yet discussed. Al-
Jayyani developed the Law of Sines, and Jamshīd al-Kāshī developed the Law of Cosines.
Through a brief stop of several centuries in China, trigonometry made its way to Europe into the
hands of such mathematicians as Isaac Newton and Leonhard Euler (Pronounced “Oiler”), who was
the first to embrace the well known modern-day concept of “abbreviating” in mathematics, creating
abbreviations for all six trig ratios that were terribly close to what we use today: sin, cos, tan, cot, sec,
and csc.
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Chapter
IV.A
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Chapter
IV.A Chapter
Overview
‣ IV.A.1 – Special Trig Ratios & the Hand Trick
This section introduces the other three trig ratios, defines the unit circle, and utilizes a “trick” for
calculating special ratios literally by hand.
‣ IV.A.2 – Special Trig Ratios of Co-terminal Angles
This section explores the concept of co-terminal angles, both positive and negative.
‣ IV.A.3 – Radian Conversions
This section investigates the origins of the alternate and much more mathematical angle
measurement unit, the radian, and allows for conversion using dimensional analysis between
degrees and radians.
‣ IV.A.4 – Special Trig Ratios of Radian Measurements
This section switches out the degrees for radians in a similar section to VI.A.2.
‣ IV.A.5 – Arc Length, Sector Area, & Rotational Motion
This section demonstrates the application and mathematical genius of radian measure in a
geometric sense as well as showing its benefits to solve physics-type problems.
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Section
IV.A.1
Special Trig Ratios &
the Hand Trick
In this section, we will discuss the unit circle, the special angles we discussed in III.A.3, and then how
to remember them with a simple trick involving counting on the fingers. My, how far we’ve come… In
chapter III.A, we stressed the close relationship between right triangles and circles, and we solved
right triangles by first getting triangles into standard form and making the hypotenuse 1. All of that
was intentionally set up for this chapter.
The Unit Circle
The unit circle has its center at the origin and a radius equal to one.
From III.A.1, we know the equation for the unit circle will be
x + y = 1. Additionally, all four of the triangles shown are,
technically, in standard position with their horizontal side being cos θ
and their vertical side being sin θ. Therefore, the point at which each
triangle touches the circle has a coordinate of (cos θ , sin θ), and the
slope of each hypotenuse is tan θ =
, which results in the following
table:
Quadrant Angles cos θ sin θ tan θ
I 0° < θ < 90° + + +
II 90° < θ < 180° - + -
III 180° < θ < 270° - - +
IV 270° < θ < 360° + - -
The Quadrangles
Quadrangles are the angles that correspond to the axes (0°, 90°, 180°, 270°, and 360°). Remember,
in the unit circle the coordinates of the circle correspond to the cosines and sines of the angles that
intersects them. The circle crosses the positive x-axis at (1,0), cos 0° = 1, sin 0° = 0, and, since the
°
slope of a horizontal line is 0, tan 0° = = 0. Likewise, since the circle crosses the positive y-axis
°
at (0,1), cos 90° = 0, sin 90° = 1, and tan 90° =
°
° =
= undefined. Furthermore, we may
remember that a vertical line has an undefined slope. We can utilize this same strategy to complete
the quadrangles:
θ cos θ sin θ tan θ
0° 1 0 0
90° 0 1 Undefined
180° -1 0 0
270° 0 -1 Undefined
360° 1 0 0
As we can see the quadrangles are fairly easy to understand. The coordinates give us the cosine
and sine, and the tangent is derived directly from those two values. However, these 5 angles are not
nearly enough to truly understand the power of the unit circle, because, simply put, the unit circle is a
tool that will lead to these trig ratios becoming trig functions in IV.B.
180
1
0.8
0.6
0.4
0.2
-1 -0.5 0.5 1
360
1
-0.2
-0.4
-0.6
-0.8
270
-1
90
1
1
0
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The Hand Trick
If we hold our left hand out with our palms facing us, we have a fairly decent approximation of the first
quadrant with our thumb at 90° and our pinky at 0°, but what values can we assign to the other three
fingers? Back in III.A.3, we discussed special triangles, and in doing so, we worked extensively with
30°, 45°, and 60°. These are the other three fingers. Firstly, referring back to III.A.3, we know:
θ cos θ sin θ tan θ
30° √
45° √
60°
√
√
= √
√
Unless we’re really good at memorization, we’ll have forgotten these by time we get to the
assignment. However, we still have our hand raised in front of our face. What’s up with that? We’ll
get to it in a second. But first, notice in the table how each of the cosine and sine values have 2s in
the denominator. This is extremely important and the only thing we must remember.
1
√3
Now to the hand… One word of warning: even if a student loves math and
loves this hand trick, it’s a bad idea to get these numbers permanently
tattooed on the tips of his/her fingers. Now that our hands are tired from
holding them up for so long, here are the steps to the hand trick.
1. Fold down the finger corresponding to the angle in the problem.
2. The fingers above the “gap” correspond to cosine, and the fingers
below correspond to sine.
3. Square root the fingers.
4. Divide by 2.
Example 1
sin 60° =
90°
60°
45°
30°
0°
We fold down our first finger as shown in the picture. Since, we are looking for sine,
we count the fingers below the gap. There are three. Therefore we square root three
and divide by two, giving us sin 60° = √
.
Example 2
tan 30° =
We fold down our fourth finger as shown in the picture. We now know cos 30° = √
and sin 30° = √
√
. Therefore, tan 30° =
. Notice the 2s cancelled out.
√
= √
√ =
√ = √
An alternative method is after folding down the fourth finger, flip the hand over. We
now have 1 finger on top and 3 on bottom, or √1 over √3.
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The Other Trig Ratios
There are three other trig ratios we must discuss: secant, cosecant, and cotangent, which are
abbreviated as sec, csc, and cot, respectively. These are reciprocal ratios of cos, sin, and tan,
respectively. In other words: sec θ =
, csc θ =
, and cot θ =
. There’s really not much
more to them.
Note: Every pair of reciprocal ratios has one “co” in it. Many students assume that the reciprocal of
cosine is cosecant, but if we remember the one co rule, we won’t make that mistake.
Example 3
sec 45° =
Secant is the reciprocal of cosine, so first we will find cosine with the hand trick. Folding down the
middle finger, we find two fingers above, so cos 45° = √
. Then sec 45° = = √
= √2.
√
Reference Angles
At this point, we may be asking ourselves “How do we find the trig ratios that
are not in the first quadrant?” The answer comes in the form of reference
angles. Reference angles are always on the interval [0°, 90°], so our hand
trick will work every time. So what are reference angles? Angles,θ, in
standard position have an initial side at the positive x-axis and a terminal side
somewhere on the unit circle in the interval [0°, 360°]. Reference angles, θ ,
are the acute angles created by the terminal side of the angle and the x-axis.
180
0.8
0.6
0.4
0.2
-1 -0.5 0.5 1
ref
-0.2
-0.4
1
-0.6
-0.8
270
-1
90
0
360
Example 4
cos 150° =
We know that 150° is in the second quadrant, and we know the cosine is the x-value, which is
negative in the second quadrant. Therefore we know our answer has to be negative. We also know
that 150° is 30° from the x-axis, so θ = 30°. So, we use the hand trick and find cos 30° = √
, which
means cos 150° = − √
.
Example 5
cot 225° =
225° is 45° from the x-axis, and is in the third quadrant, which has positive slopes. Therefore, we
know cot 225° = cot 45°. We find tan 45° = 1 by the hand trick or table, so cot 45° =
= = 1.
°
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IV.A.1 Independent Practice
Calculate the following without a calculator.
1. sin 0°
2. cos 0°
3. tan 0°
4. csc 0°
5. sec 0°
6. cot 0°
7. sin 30°
8. cos 30°
9. tan 30°
10. csc 30°
11. sec 30°
12. cot 30°
13. sin 45°
14. cos 45°
15. tan 45°
16. csc 45°
17. sec 45°
18. cot 45°
19. sin 60°
20. cos 60°
21. tan 60°
22. csc 60°
23. sec 60°
24. cot 60°
25. sin 90°
26. cos 90°
27. tan 90°
28. csc 90°
29. sec 90°
30. cot 90°
31. sin 330°
32. cos 270°
33. tan 210°
34. csc 150°
35. sec 300°
36. cot 180°
37. sin 240°
38. cos 135°
39. tan 315°
40. sec 225°
41. csc 120°
42. cot 360°
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Section
IV.A.2
Special Trig Ratios of
Co-Terminal Angles
In III.B.1, we discussed the concept of negative angles, which rotate clockwise from the positive x-
axis, but what wasn’t mentioned were angles larger than 360°. How do we find trig ratios for those?
The answer is co-terminal angles.
Co-Terminal Angles
We already know that on the unit circle, 0° is equivalent to 360°, and they have identical trig ratios.
Co-terminal angles have the same terminal side and are 360° apart. Therefore 20°, 380°, and -340°
are all co-terminal because there’s 360° difference between 20° and 380°, and there’s 360° difference
between 20° and -340°. -340° and 380° are co-terminal because they are 720° apart, 720 being a
multiple of 360. We will need to convert all angles to their co-terminal equivalent between 0° and
360°.
Example 1
Calculate the co-terminal angle of 2475° on the interval [0°, 360°).
We can either subtract 360° repeatedly from 2475° until we have an angle between 0° and 360°. The
other option is pretty nifty. We divide
= 6.875, subtract off the whole number (this is how many
360s there are between 2475° and its co-terminal angle), and then multiply by 360°. The co-terminal
angle is . 875(360°) = 315°.
Example 2
sin(−750°) =
First, we calculate the co-terminal angle: −
= −2.083 . We will have to add 3, to get a positive
decimal value: −2.083 + 3 = .916 and . 916(360°) = 330°. So sin(−750°) = sin 330° = − sin 30° = − .
Example 3
csc(4080°) =
= 11. 3 (11. 3 − 11)(360°) = 120°.
sin 4080° = sin 120° = sin 60° = √
.
csc 4080° =
√ = √
.
This section’s exercises are similar to the previous section, but understanding these special angles
well will allow for a simple transition in the following sections and chapters.
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IV.A.2 Independent Practice
Calculate the exact values of the following.
1. sin 390°
13. sin 1980°
2. cos 720°
14. cos(−660°)
3. tan 585°
15. tan 1230°
4. csc 510°
16. csc 2790°
5. sec 675°
17. sec 1770°
6. cot 450°
18. cot(−2565°)
7. sin 1290°
19. sin 4455°
8. cos(−1020°)
20. cos 2100°
9. tan 3360°
21. tan(−3555°)
10. csc(−1665°)
22. csc 900°
11. sec 2820°
23. sec(−2310°)
12. cot(−675°)
24. cot(−405°)
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Section
IV.A.3 Radian
Conversions
Multiple ancient cultures worshipped and studied the movement of the stars. The ancient
Babylonians and other societies from long ago approximated the year to be about 360 days long.
They weren’t too far off. They observed the stars circling the North Star seemed to move
of a
circle each night, and this is more than likely where the degree came from, being
of a circle. 360
would have been a perfect number as it divides evenly by 4 (seasons) and 12 (months), but we do
have a pesky 5.25 days unaccounted for. All in all, it’s impressive that something that started out as
an error lasted so long. Likewise, due to its less than mathematical nature, advanced mathematics
can’t use degrees, so we must change our point of view.
Radians
A radian is the measurement of an arc of a circle and, by consequence, the angle it subtends.
Luckily, a radian can be considered a linear measurement (since it measures distance), so advanced
mathematics is compatible. Since we’re talking about the circle itself and not an angle inside of it,
we’re more concerned with circumference. We know from geometry C = 2πr, where r is the radius.
What this says is that we can wrap the length of the radius around the outside of a circle 2π
(approximately 6.28) times, regardless of the actual numerical length. Therefore, we turn the straight
radius into a curved radian, of the same length, and since the measurement is now in terms of “1”
radius, we have essentially formed a unit circle.
We have just shown that there are 2π radians in one revolution of a circle, therefore 2π radians is
equivalent to 360°. Half of a revolution is therefore π radians or 180°, and a quarter revolution is
radians or 90°. Understanding this general conversion is the key to converting between radians and
degrees.
Example 1
Convert 165° to radians.
°
°
Example 2
=
radians. Here we can see the degrees cancel out.
Convert radians to degrees.
°
°
= 12°.
Example 3
= 12°. Here we see the π rad cancel out. One other option is to replace π with 180°.
Convert 3 radians to degrees.
°
= 171.887°. In this case, we cannot cancel out the π, nor is there a π in which we can
substitute 180°. It is bad form to leave π in the denominator.
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IV.A.3 Independent Practice
Convert the following into radians.
1. 80°
7. 900°
2. 25°
8. 400°
3. 100°
9. 555°
4. 36°
10. −120°
5. 210°
11. −315°
6. 320°
12. −250°
Convert the following to degrees.
13.
19.
14.
20.
15.
21. −
16.
22. 12
17.
23. −4
18.
24. 23
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Section
IV.A.4
Special Trig Ratios of
Radian Measurements
We are going to rehash IV.A.2 once again, but this time, we will be finding trig ratios of radian
measurements. However, since we’ve become accustomed to finding them with degree
measurements, we’re not going to throw degrees completely out the window. The good news is that
the more mathematical radian makes a couple things a little easier.
Co-Terminal and Reference Angles
If we start at the positive x-axis (0 radians) and make one revolution, we end at 2π radians, and if we
make another revolution, we end at 4π radians. Not surprisingly, two angles are co-terminal if they
are a multiple of 2π radians apart. Therefore π radians and 3π radians are co-terminal. In fact any
even multiple of π is co-terminal with 0, and any odd multiple of π is co-terminal with π.
Because radians are often represented as fractions of π, reference angles become really easy to find.
For instance, has a reference angle of because it’s of the way from 0 to π and has of the way
to go to reach the x-axis. Likewise,
will have a reference angle of . In fact, any angle will
have a reference angle of .
Example 1
sin
=
Dividing
, we get 4π with π left over. The 4π disappears because it’s equivalent to 0.
So, sin
= sin 4π + = sin √
= sin ° = sin(60°) = , according to the hand trick.
Example 2
sec
=
Let’s begin by calculating the cosine:
cos
= cos 21π + = cos π + = − cos
√
= − cos 30° = − . So, sec = −
= − √
.
√
has a reference angle of , and because it’s equivalent to π + , we know it’s in the third quadrant
(half a revolution plus a bit counter-clockwise) so its cosine is negative.
Example 3
sin −
=
sin −
= sin −6π − = sin −7π + = sin π + = − sin √
= − sin 45° = − .
−6π − = −7π + because the second term must be positive. We subtracted π from the first term
and added it to the second term. π + is in the fourth quadrant (where sine is negative) because
we’ve gone half a revolution and then almost another half.
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IV.A.4 Independent Practice
Calculate the exact values of the following trigonometric ratios.
1. sin
13. sin
2. cos 14. cos 7π
3. tan 15. tan
4. csc 16. csc
5. sec
17. sec
6. cot
18. cot
7. sin −
19. sin −
8. cos −
20. cos −
9. tan −
21. tan −
10. csc (−6π)
22. csc −
11. sec −
23. sec −
12. cot −
24. cot −
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Section
IV.A.5
Arc Length,
Sector Area, and
Rotational Motion
In this section, we are going to apply the information from the chapter to solve real-world problems.
Plus, we want to see what’s just so gosh-darn important about radians.
Arc Length
We already know that a radian is a unit used to find arc measure. An arc of measure 3 radians
indicates that we can take the radius of the circle and wrap three of them along the edge of the circle.
So if we knew the numerical length of the radius, all we would have to do to find the numerical length
of the arc would be to multiply the radius length by 3. This revelation leads to the fact:
Example 1
l = rθ, where r is the length of the radius, and θ is the arc measure in radians
In a circle of radius 8 cm, what is the length of the arc subtended by a central angle of 65°?
We must first convert 65° to radians.
°
°
radius.
rad (8 cm/rad) =
radian and radius cancel, as do the degrees.
Sector Area
=
radians. We then multiply this value by 8 cm per
cm. We could calculate it all at once:
°
°
=
cm. The
Sector area is a little trickier since it is a portion of the area of the circle. The area of a circle is
A = πr , and the area of a sector is A =
° πr = θ
° r = θ
r , due to degree to radian
conversion.
A = θr , where θ is in radians
With radians we don’t have to worry about figuring out percentages of a circle or any such nonsense.
They take care of that for us.
Example 2
Calculate the area of the sector of a circle of radius 5 inches and a central angle of 125°.
A =
°
°
=
= 27.271 in
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Rotational Motion
Rotational Velocity is a natural byproduct of arc length and arc measure. As we may or may not be
aware, the velocity of an object is equal to displacement (distance with a direction) over time, which
means the change in arc length (a linear measurement) over time results in linear velocity, v.
Likewise, the change in arc measure (radians) over time results in angular velocity, ω. We may
also notice that v = rω, which is fairly significant.
Example 3
Three bugs are sitting on a record spinning at 20 rpm, one at the center, the next 5 cm from the
center, and the last 10 cm from the center. What are the bugs’ linear velocities?
All three bugs have the same angular velocity:
Bug A:
Bug B:
Bug C:
.
.
.
= 0 cm/min
= 200π cm /min
= 400π cm/min
.
= 40π rad/min.
The trick to doing advanced problems is understanding that objects that share the same axle (center),
have the same angular velocity, and objects that connect by their circumferences share the same
linear velocity. For instance in the following diagram:
d
b
a
c
e
Wheels a and b share the same angular velocity, a and c share the same linear velocity, c and d
share the same angular velocity (different from a’s and b’s), and d and e share the same linear
velocity.
Example 4
If a cycler spins his pedals, and thus a 8 inch diameter gear, at 20 rpm, that gear is connected by a
chain to a 6 inch diameter gear, which is connected directly to the 36 inch diameter wheel, how fast is
the bicycle moving in miles per hour?
The first and second gears share the same linear velocity, and the second gear and the wheel share
the same angular velocity. So, first we find the angular velocity in radians per minute of the first gear,
then calculate the gear’s linear velocity (shared by the second gear). We calculate the second gear’s
angular velocity (shared by the wheel), and then, we calculate the linear velocity of the wheel.
Finally, we have to convert the units from inches per min to miles per hour.
= 960π inch/min. The use of 4, 3, and 18, was because we need the
.
radii of the three rotating objects, which were half the given diameters.
.
= 2.856 mph, which is not terribly impressive.
.
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IV.A.5 Independent Practice
Determine arc lengths and sector areas given the radii and central angles of the following
circles.
1. r = 5 cm, θ = 30°
2. r = 8 in, θ = 45°
3. r = 2 ft, θ = 15°
4. r = 4 yd, θ = 120°
5. r = 3 mi, θ = 90°
6. r = 10 mm, θ = 135°
7. r = 6 m, θ = 100°
8. r = 12 cm, θ = 200°
9. r = 15 ft, θ = 240°
10. r = 20 mi, θ = 225°
11. r = 9 mm, θ = 270°
12. r = 1 in, θ = 300°
Answer the following rotational motion questions.
13. A wheel has a radius of 8 cm and an angular velocity of 10 radians per second. What is the
linear velocity on its circumference? At a point halfway from the axle to the circumference? At
the axle?
14. A 12-inch-diameter gear’s teeth are turning at 4 inches per second. What is the angular
velocity of the teeth? At a point halfway from the axle to the circumference? At the axle?
15. A cog of radius 9 mm is rotating at 30 rpm. What is the linear velocity of the cog’s teeth?
16. A 20-inch-radius wheel is travelling at 50 mph. What is its angular velocity in rpm?
17. Two wheels are connected by a belt. The 2-foot-radius wheel is rotating at 6 radians per
second. What is the angular velocity of the 3-foot-radius wheel?
18. The teeth of gear A (diameter 6 inches) are interwoven with the teeth of gear B (diameter 4
inches) which are also interwoven with the teeth of gear C (diameter 10 inches). If gear A is
rotating at 25 rpm, what are the rpm of the other two gears?
19. Two gears, one with a radius of 7 mm and the other with a radius of 12 mm, are glued together
so that they share an axle. If the teeth of the larger gear are being rotated at 30 mm per
second, what is the linear velocity of the teeth of the smaller gear?
20. If the smaller gear from number 19 is connected via belt to a third gear of radius 10 mm, what
is the angular velocity of the third gear in rpm?
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Chapter
IV.B
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Chapter
IV.B Chapter
Overview
‣ IV.B.1 – Graphing Sine & Cosine
This section will develop the concept of trigonometric functions through the graphing of the two
most familiar functions. We will also investigate function transformations and their effects on
amplitude, period length, vertical shift, and phase shift.
‣ IV.B.2 – Graphing Secant, Cosecant, Tangent, &
Cotangent
This section will utilize the knowledge of sine and cosine graphs to synthesize the notion of
secant, cosecant, tangent, and cotangent graph, as well as explore their functional
transformations.
‣ IV.B.3 – Inverse Trig Functions & Initial Values
This section will investigate inverse trig functions, and define the notion of restricted domains.
‣ IV.B.4 – Arcsine, Arccosine, & Arctangent
This section will explore the notions of continuous inverse operations and the calculation of all
possible radian values.
‣ IV.B.5 – Solving Simple Trigonometric Functions
This section will allow for the solving of trig equations for angular values.
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Section
IV.B.1 Graphing
Sine & Cosine
So far, we’ve been slightly wishy-washy with our uses of the words “trig ratios” and “trig functions”.
Sine and cosine (and the rest) are functions; sin(θ) and cos(θ) are ratios, which are the outputs of
the sine and cosine functions (angles being the inputs). We will now need to tweak a couple things,
however: θ will now indicate the use of degrees and x will indicate the use of radians. The question
many may be wondering is “Why?”
Although we have placed the unit circle on a coordinate grid, it should not be considered the typical
Cartesian, (x,y), coordinate system. The circle itself is not a function or a graph per se; it is, in fact, a
device to assist in the understanding of trigonometric ratios. The graphs of sine and cosine are much
more fascinating.
Sine Graphs
As we know, the sine of an angle is the vertical component of the unit circle at that angle value.
Additionally, we can revolve counter-clockwise around the unit circle indefinitely, and we know that
the sine ratios (vertical lengths) repeat every 2π radians (x-values), which means the sine function will
also repeat. We refer to these types of repeated functions as periodic functions, one repetition being
a single period. We also know that if we go clockwise in the unit circle, we can still find sine ratios of
the negative angles, which means the domain of a sine function is indeed all real numbers (R). So
what’s the range? Well, how high and low can we go in a circle with a radius of 1? The minimum
possible vertical displacement is -1, and the highest is 1. Therefore the range of a sine function is
[−1,1]. Furthermore, we can probably guess that because a circle is one big curve, the graph of the
sine function will also have curves. We sure know a lot for not ever having seen the graph of
f(x) = sin(x) before… Let’s look at the values for the first 2π radians:
x 0
sin x 0
√
√
1 √
√
π
0 −
− √
− √
−1
− √
− √
−
2π
0
Plotting these points:
y
x
And connecting the points:
y
y
x
x
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Every sine function graphs in the same way. It starts at the sinusoidal axis and is broken into four
parts (that coincide to the four quadrants), achieving a maximum at , the sinusoidal axis again at , a
minimum at , and the sinusoidal axis again at the end of the period. It repeats in both directions,
periodically, like this infinitely.
Example 1
Graph f(x) = sin x − + 5.
Because sine is a function, transformations affect as they do other functions. The 5 will translate the
function up 5, so the sinusoidal function is now at 5. Because the
inner transformations affect a function oppositely, the function will
6
then undergo a phase (horizontal) shift right . This translates the
origin of the sine function from (0,0) to ( 5
, 5). The graph, however
−
remains the same, otherwise. The amplitude is still 1, so we add 1 to
4
and subtract 1 from 5 to get a maximum of 5 and a minimum of 4.
The period is still 2π, so we add and subtract 2π to get − and
.
Example 2
Graph f(x) = 3 sin x.
3
4π
The 3 will dilate the function vertically by 3, increasing the maximum to
3 and decreasing the minimum to -3. The causes a horizontal
dilation by a scale factor of its reciprocal, 2, which means we multiply
2(2π), increasing the period length to 4π.
0
−4π
−3
0
Example 3
Graph f(x) = −5 sin x + 2π + 3.
First, we must factor out as we did in II.B.4 and II.B.5, resulting in
−5 sin (x + 12) + 3.
The origin shifts from (0,0) to (-12,3).
The period becomes (2π) = 12. Add 12 to and subtract 12 from -12.
The amplitude becomes 5. Add 5 to and subtract 5 from 3.
The entire function reflects across the sinusoidal axis.
3
−24
8
−2
-12
0
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Cosine Graphs
The only difference between sine graphs and cosine graphs is the initial shape of the graph. Cosine
graphs transform in the same way as sine graphs, but they start out differently. Remember, the
cosine of an angle is the horizontal component of the unit circle. Let’s create a table and a graph of
cos x the same we did for sin x.
x 0
cos x 1 √
√
0 −
− √
− √
π
−1
− √
− √
−
0
√
√
2π
1
As we can see, the cosine graph starts at the maximum, intersects the sinusoidal axis at of the
period, achieves a minimum at of the period, intersects the sinusoidal axis again at of the period,
and returns to the maximum at the end of the period.
Example 4
Graph f(x) = 7 cos(2x − 8) − 4.
3
First, we factor out 2, 7 cos2(x − 4) − 4.
The origin transforms to (4, −4).
The amplitude increases to 7.
The period decreases to (2π) = π.
−4
4 − π
−11
4
4 + π
Example 5
Graph f(x) = 7 cos(−2x − π) + 3.
7 cos −2 x + + 3.
The origin moves to − , 3.
The amplitude increases to 7.
The period decreases to (2π) = π.
3
−
10
−4
−
π
The graph reflects across the x = − , although when
cosine reflects horizontally, it remains the same.
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IV.B.1 Independent Practice
Label all tick marks based on the given equation.
1. y = 5 sin(x) + 3
5. y = cos(3x + 9)
2. y = sinπ(x − 3)
6. y = 5 cos (x − 6) + 10
3. y = 3 sin(4x)
7. y = 3 cos(2x − 10) + 2
4. y = sin(x + 4) − 6
8. y = 5 cos x + 3π − 7
Determine a sine and cosine equation for the following graphs.
10
10
5
-4
8
4
-13 -1
9.
0
2
11.
-2
-7
1
7
-2
-5
2
5
- 6
3
2
10.
3
2
12.
6
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Section
IV.B.2
Graphing
Secant, Cosecant,
Tangent, & Cotangent
We spent a long time discussing and describing the graphs of sine and cosine. We did so, because
we will use that base knowledge to define the graphs of the other four trigonometric functions.
Secant & Cosecant Graphs
As long as we understand how to graph sine and cosine functions, the concept behind secant and
cosecant ratios, and general knowledge of reciprocals, secant and cosecant pretty much graph
themselves. The reciprocal of 0 is which is undefined by the First Mathematical Commandment,
and results in a vertical asymptote. The reciprocal of 1 is still 1, so the maximums and minimums of
sine and cosine stay the same. (Fig. 1) As a number gets smaller, its reciprocal gets bigger, and
when a positive number gets infinitely close to 0, its reciprocal approaches infinity, and when a
negative number gets infinitely close to 0, its reciprocal approaches negative infinity, resulting in the
bold graphs of cosecant and secant (Fig. 2).
y
y
sin(x)
csc(x)
x
x
Fig. 1
Fig. 2
cos(x)
y
sec(x)
y
x
x
Example 1
Graph f(x) = 5 sec (x − 8) − 2.
The origin translates to (8, −2).
The “amplitude” increases to 5.
The period increases to (2π) = 8.
Example 2
Graph f(x) = −3 csc(3x + ) + 5.
−2
0
3
−7 16
8
f(x) = −3 csc 3 x + + 5.
The origin translates to , 5.
The “amplitude” increases to 3.
The period decreases to (2π) = .
The graph reflects vertically.
5
−
8
2
−
Note: Secant and cosecant graphs don’t have amplitudes per se, hence the quotations. Instead, we
have to think of the amplitudes applying to their reciprocals, sine and cosine.
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Tangent & Cotangent Graphs
Tangent and cotangent graphs are unique in comparison to the others. Their uniqueness comes
about because they are not a specific value on the unit circle, but because they are slopes (or
reciprocals of slopes) that are formed from two values on the unit circle. Remember, the slopes at 0
and π are 0, and the slopes at and are undefined (vertical asymptotes). Additionally we know
tan = 1, tan = −1, tan = 1, and tan = −1. We also may remember from our study of
vectors in III.B.1 that the tangent values of two angles 180° (π rad) apart are equal, which means the
period of tangent is only π, not 2π like the others. Following the trends of the points, we get:
y
x
Notice, we have twice as many periods in the same domain as the other four trig functions, because
the period is half as long.
The cotangent graph can be derived from its reciprocal relationship with tangent. Remember that the
reciprocal of 0 is undefined, and, conversely, the reciprocal of undefined is 0. The reciprocal of 1 is 1,
and the reciprocal of -1 is -1. If we plot these facts on the coordinate grid, connect the dots, and
follow the trends, we will get a cotangent graph.
y
x
Example 1
Graph f(x) = 3 tan (x + 6) − 8.
The origin moves to (−6, −8).
The “amplitude” dilates to 3, and the period dilates to (π) = 5.
−8 −11
−5
−11
−6
−1
Example 2
Graph f(x) = 2 cot(−5x + 2π) + 1.
2 cot −5 x − + 1.
1
3
−1
The origin moves to , 1.
The “amplitude” dilates to 2, and the period dilates to (π) = .
The graph reflects horizontally.
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IV.B.2 Independent Practice
Label all tick marks based on the following equations.
1. y = 3 sec (x + 2) − 5
7. y = 5 tan(3x + 9π) − 2
8. y = 3 tan (x − 2) − 8
2. y = 5 sec 3 x + + 2
9. y = 9 tan(3πx − 12π) − 4
3. y = 8 sec2π(x − 9) + 3
10. y = 6 cot 6 x + + 2
4. y = 2 csc(4x − π) + 8
5. y = 5 csc x + − 2
11. y = 12 cot (x − 5) + 9
12. y = 2 cot(5x − 10π) − 5
6. y = 7 csc (x − 5) − 9
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Section
IV.B.3
Inverse Trig Functions &
Initial Values
Because the trig functions are periodic, they fail the horizontal line test, making them non-invertible,
however, if we restrict the domain of the original functions, we can make them invertible.
Inverse Sine
sin(x) (D: R; R: {[−1,1]}) sin(x) (D: − , ; R: {[−1,1]}) sin (x) (D: {[−1,1]}; R: − , )
y
y
y
x
x
x
Inverse Cosine
cos(x) (D: R; R: {[−1,1]}) cos(x) (D: {[0, π]}; R: {[−1,1]}) cos (x) (D: {[−1,1]}; R: {[0, π]})
y
y
y
x
x
x
Inverse Cosecant
csc(x) (D: {x ≠ aπ}; R: {(−∞, −1] ∪ [1, ∞)})
csc(x) (D: − , 0 ∪ 0, ; R: {(−∞, −1] ∪ [1, ∞)}) csc (x) (D: {(−∞, −1] ∪ [1, ∞)}; R: − , 0 ∪ 0, )
y
y
y
x
x
x
Inverse Secant
sec(x) (D: x ≠ ; R: {(−∞, −1] ∪ [1, ∞)}) sec(x)
(D: 0, ∪ , π ; R: {(−∞, −1] ∪ [1, ∞)}) sec (x) (D: {(−∞, −1] ∪ [1, ∞)}; R: 0, ∪ , π)
y
y
y
x
x
x
Inverse Tangent
tan(x) (D: x ≠ ; R: R) tan(x) (D: − , ; R: R) tan (x) (D: R; R: − ,
y
y
y
x
x
x
Inverse Cotangent
cot(x) (D: {x ≠ aπ}; R: R) cot(x) (D: (0, π); R: R) cot (x) (D: R; R: (0, π))
y
x
y
x
y
x
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This probably seems like a lot to take in. Here are the real keys to inverse trig functions.
The angles sin x , csc x, and tan x are restricted to quadrants I and IV of the unit circle.
The angles cos x , sec x, and cot x are restricted to quadrants I and II of the unit circle.
Since inverse trig functions take in a ratio and produce an angle, we can utilize the hand trick to
perform the reverse of how we’ve used it before.
Example 1
sin =
is the sine value (vertical component of the unit circle), so we have to fold
down the fourth finger in order to get a sine value of . The fourth finger
represents 30°, so sin = .
Example 2
sec −√2 =
Since the secant value, −√2, is negative, the angle is in the second quadrant of the
unit circle where the horizontal components are negative. Since sec x = −√2,
cos x = −
= − √ . Holding down the middle finger produces the necessary two
√
fingers for the cosine value, which means the reference angle is 45° or . Since the actual angle is in
the second quadrant, the correct angle is either 135° (45° from 180°) or
( from π).
Example 3
tan − √
=
Since the tan x = − √ , x is in the fourth quadrant of the unit circle, where the slopes
are negative. − √ = −
, which means sin x = − √
hold down the fourth finger, giving us a reference angle of 30° or
angle (in the fourth quadrant) is either −30° or − .
√
and cos x = , which means we
. The correct
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IV.B.3 Independent Practice
Solve the following in degrees.
1. sin √
=
7. sin (0) =
2. cos (−1) =
8. cos − =
3. tan √3 =
9. tan (0) =
4. sec (2) =
10. sec −√2 =
5. csc − √
=
11. csc ( 2) =
6. cot (1) =
12. cot − √
=
Solve the following in radians.
13. sin (−1) =
19. sin − √
=
14. cos √
=
20. cos (−1) =
15. tan (−1) =
21. tan −√3 =
16. sec (1) =
17. csc (−2) =
22. sec √
=
23. csc (1) =
18. cot (0) =
24. cot √3 =
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Section
IV.B.4
Arcsine, Arccosine, &
Arctangent
Arcsine, arccosine, and arctangent are the actual non-function inverses of sine, cosine, and tangent
without the restricted domain. These calculate all angles that result in the given ratios, but to
understand them, we will have to look back at the unit circle.
Arccosine
y
The cosine value, as we’re well aware of by now, is the horizontal
component of the unit circle. The two angles shown have identical cosine
values, and, likewise, have identical reference angles. In this case ±x
would be the two angle values. Therefore:
x
-x
1
cos(x)
1
x
arccos(x) = ±cos x + 2πn
arccos(x) = ± cos x + 360°n
Arcsine
The sine value is the vertical component of the unit circle. The two angles
shown have identical sine values and identical reference angles.
Remembering the angle is measured from the positive x-axis, we can
show that the two angles shown add to π. Therefore the two angles are x
and π − x. Thus:
sin(pi-x)
y
1
pi-x
x
x
1
sin(x)
x
arcsin(x) =
sin x + 2πn
π − sin x + 2πn
arcsin(x) =
sin x + 360°n
180° − sin x + 360°n
Arctangent
The tangent value is the slope of the radius of the unit circle, and, as we’ve shown before, the period
length of tan(x) is π. Based on this information, and the shape of the tangent graph, there are no
y
shared tangent values in a single period. So:
arctan(x) = tan x + πn
arctan(x) = tan x + 180°n
x
Understanding these facts is the key to learning how to use them.
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Example 1
arcsin √
=
sin √ + 2πn
=
+ 2πn
π − sin √ + 2πn
+ 2πn
or
sin √
+ 360°n 60° + 360°n
=
180° − sin √ + 360°n 120° + 360°n
Example 2
arccos − √
=
± cos − √
+ 2πn = ± + 2πn
or
± cos − √ + 360°n = ±135° + 360°n
Example 3
arctan(−1) =
tan (−1) + πn = − + πn
or
tan (−1) + 180°n = −45° + 180°n
Example 4
arcsin − =
sin − + 2πn
π − sin − + 2πn = − + 2πn
+ 2πn
or
sin −
+ 360°n
+ 360°
180° − sin − = −30°
+ 360°n 210° + 360°
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IV.B.4 Independent Practice
Determine all values for the following arc-functions in degrees.
1. arcsin √ =
5. arccos 1 =
2. arccos − =
6. arctan(−1) =
7. arcsin 1 =
3. arctan 0 =
8. arccos √ =
4. arcsin − √ =
Determine all values for the following arc-functions in radians.
9. arcsin 0 =
13. arccos 0 =
10. arccos √
=
14. arctan−√3 =
11. arctan √
=
15. arcsin − =
12. arcsin(−1) =
16. arccos(−1) =
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Section
IV.B.5
Solving Simple
Trigonometric Functions
We will now combine all knowledge of trigonometric functions and functions in general to solve for
specific angular values in both degrees and radians.
Example 1
Solve for the first four positive values of θ: 4 sin2(θ − 5) + 7 = 9.
4 sin2(θ − 5) = 2
sin2(θ − 5) =
2(θ − 5) = arcsin =
sin + 360°n
30° + 360°n
=
180° − sin + 360°n 150° + 360°n
15° + 180°n
θ − 5 =
75° + 180°n
20° + 180°n
θ =
80° + 180°n
The first four positive values are {20°, 80°, 200°, 260°}.
Example 2
Solve for the first four positive values of x: 6 cos (x + 2) = −3√3.
cos √
(x + 2) = −
(x + 2) = arccos −
√
= ± cos − √
+ 2πn = ± + 2πn
x + 2 = 12
π ± 5π 6
+ 2πn = ±10 + 24n
x = 8 + 24n
−12 + 24n
The first four positive values are {8,12,32,36}.
Example 3
Solve for the first four positive values of x: 5 tan 2 x − − 8 = −13.
tan 2 x − = −1
2 x − = arctan(−1) = tan (−1) + πn = − + πn
x = + − + πn = − + n = + n
, ,
,
.
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IV.B.5 Independent Practice
Solve the following problems for the first four positive degree values.
1. 5 sin3(θ − 23°) + 7 = 12
4. −2 sin (θ + 85°) + 5 = 6
3. 4 tan (θ − 22°) + 7 = 3
2. 2 cos5(θ + 76°) = √3
5. 8 cos5(θ − 120°) − 5 = −5
6. √3 tan (θ + 95°) + 8 = 7
Solve the following problems for the first four positive radian values.
7. 6 sin (x − 4) + 2 = 5
11. 5 cos
(x − 6) + 9 = 4
8. 7 cos (x + 7) − 4 = 3
12. 7 tan 2π x + + 8 = 8
9. 2 tan (x − 9) + 2√3 = 4√3
13. 4 sin 2 x + − 3 = −1
10. 8 sinπ(x + 3) + 8√2 = 4√2
14. 8 tan x − − 3 = 5
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Chapter
IV.C
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Chapter
IV.C Chapter
Overview
‣ IV.C.1 – Law of Sines
This section will provide a rule for solving AAS and ASA oblique triangles, as well as a proof of the
area formula learned in III.A.4.
‣ IV.C.2 – Law of Cosines
This section will allow for the solving of SSS and SAS oblique triangles using both forms of the
Law of Cosines.
‣ IV.C.3 – Three Dimensional Vectors
This section will investigate three dimensional vectors and their different aspects.
‣ IV.C.4 – Cross Product
This section will develop a new type of vector product that results in a vector rather than the dot
products numerical value.
‣ IV.C.5 – Cross Product Applications
This section will apply our knowledge of three-dimensional vectors and cross product to find areas
of triangles knowing only their vertices’ coordinates and determining equations for entire planes
within three dimensional space.
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Section
IV.C.1 Law of Sines
We may or may not remember the following fact about triangles: the smallest angle is opposite the
smallest side, the largest angle is opposite the largest side, and the remaining angle is opposite the
remaining side. This fact is cemented in the Law of Sines:
=
=
We will use to solve AAS and ASA triangles, i.e. triangles of which we know a
side and any two angles. Again, if know two angles, we can easily calculate the third.
Example 1 (AAS)
A
b
c
C
a
B
Given the triangle at right, calculate a, c, and C.
5
C
a
We can calculate C = 180° − 32° − 47° = 101°.
° °
Then solving = through cross multiplication leads to a = 3.623.
Solving
°
=
°
results in c = 6.711.
32
c
47
Example 2 (ASA)
Given the triangle at right, calculate a, b, and C.
C = 180° − 22° − 67° = 91°
° °
= → a = 2.623
°
=
°
→ 6.445
22
b
7
C
a
67
Prove: Area =
.
Proof:
Area = ab sin C as proved in III.A.4.
Area =
Area =
a sin C since =
and b =
.
by simplification. Q.E.D.
Note: The Law of Sines should never be used to solve an angle since sin could possibly return a
negative angle but never an obtuse angle. It might be an easy law, but it’s much too limited.
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IV.C.1 Independent Practice
Solve the following AAS and ASA triangles. Round to three decimal places.
b
57
a
b
C
a
1.
64
10
B
6.
72
7
56
b
121
a
b
C
a
2.
43
6
B
7.
30
8
68
b
143
a
b
C
a
3.
23
6
B
8.
12
5
128
b
57
a
b
C
a
4.
36
3
B
9.
48
5
75
b
42
a
b
C
a
5.
87
12
B
10.
33
1
98
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Section
IV.C.2
Law of
Cosines
As discussed in the previous section Law of Sines is very limited and shouldn’t be used to solve
angles. That’s where the Law of Cosines comes in; it’s much more versatile and can be used to
solve SAS and SSS triangles.
c = a + b − 2ab cos C
C = cos
Note: Side c is whatever side is given, and angle C is the angle opposite c. We must make sure we
calculate correctly.
Example 1
A
b
C
C
a
B
Given the triangle at right, calculate a, B, and C.
7
C
a
a = 5 + 7 − 2(5)(7) cos 35° = 16.659 → a = √16.659 = 4.082.
B = cos
()() = 120.029°.
C = 180° − 35° − B = 24.971°.
35
5
B
Note: We must make sure to store values in our calculator as we solve for them, so as not to use a
rounded value to calculate something else. In the previous example the variable a was used in
equation B and variable B was used in equation C to signify the storage of the values.
Example 2
Given the triangle at right, calculate A, B, and C.
5
C
8
A = cos
= 92.866°
()()
B = cos
= 38.625°
()()
C = 180° − A − B = 48.509°
A
6
B
The Law of Cosines can be used to prove Hero’s Formula from III.A.4, but we don’t know enough yet
to prove it. We will do so in VI.C.1.
In the mean time, let’s sit back, relax, and do some math!
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IV.C.2 Independent Practice
Solve the following SAS and SSS triangles. Round to three decimal places.
7
C
a
4
C
8
1.
52
9
B
6.
A
6
B
8
C
a
12
C
15
2.
123
4
B
7.
A
5
B
7
C
a
3
C
10
3.
78
5
B
8.
A
6
B
9
C
a
12
C
9
4.
95
2
B
9.
A
5
B
12
C
a
7
C
3
5.
23
6
B
10.
A
13
B
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Section
IV.C.3
Three Dimensional
Vectors
Technically speaking, three dimensional vectors don’t necessarily fall under the general umbrella that
is trigonometry. In fact these last three sections won’t have very much trig in them at all. Woo hoo!
Three dimensional vectors are exactly like two dimensional vectors except trying to describe the
direction in terms of an angle would drive every crazy, so we won’t worry about it. The same
formulas from III.B still work for three dimensions with some minor alterations. Given a⃗ = 〈a , a , a 〉
and b⃗ = 〈b , b , b 〉:
Example 1
|a⃗| = a + a + a
a⃗ ∙ b⃗ = a b + a b + a b
Calculate the angle, θ, between a⃗ = 〈3,4,7〉 and b⃗ = 〈2,5, −3〉 when they are placed tail-to-tail.
a⃗ ∙ b⃗ = 3(2) + 4(5) + 7(−3) = 6 + 20 − 21 = 5
|a⃗| = √3 + 4 + 7 = √9 + 16 + 49 = √74
b⃗ = 2 + 5 + (−3) = √4 + 25 + 9 = √38
a⃗ ∙ b⃗ = |a⃗|b⃗ cos θ → 5 = (74)(38) cos θ → θ = 84.590°.
Example 2
Calculate the scalar projection, p, and vector projection, p⃗, of a⃗ = 〈3,5, −1〉 onto b⃗ = 〈2, −5, −3〉.
a⃗ ∙ b⃗ = 3(2) + 5(−5) − 1(−3) = −16
|a⃗| = √9 + 25 + 1 = √35
b⃗ = √4 + 25 + 9 = √38
p = ⃗∙ = −
⃗ √
p⃗ = p ⃗
= − 〈,,〉
= 〈−
,
, 〉.
⃗ √ √
Example 3
Determine the point that is 25% from a(2,5, −1) to b(−6,1,11).
ab ⃗ = 〈−6 − 2,1 − 5,11 − (−1)〉 = 〈−8, −4,12〉
a⃗ + ab ⃗ = 〈2,5, −1〉 + .25〈−8, −4,12〉 = 〈2,5, −1〉 + 〈−2, −1,3〉 = 〈0,4,2〉
The resultant is a position vector that points to the answer, (0,4,2).
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IV.C.3 Independent Practice
Determine the following given each pair of vectors.
1. a⃗ =< 3,5,4 > & b⃗ =< 2,6, −3 >
a. 3a⃗ + 2b⃗ =
b. |a⃗| =
c. b⃗ =
d. a⃗ + b⃗ =
2. a⃗ =< 5, −2,7 > & b⃗ =< −3,8, −4 >
a. 2a⃗ − 5b⃗ =
e. a⃗ ∙ b⃗ =
f. θ =
g. p =
h. p⃗ =
e. a⃗ ∙ b⃗ =
b. |a⃗| =
f. θ =
c. b⃗ =
g. p =
d. a⃗ + b⃗ =
h. p⃗ =
3. a⃗ =< 5,2,7 > & b⃗ =< −2, −5,9 >
a. a⃗ − 2b⃗
b. |a⃗| =
c. b⃗ =
d. a⃗ + b⃗ =
e. a⃗ ∙ b⃗ =
f. θ =
g. p =
h. p⃗ =
4. Find the displacement vector from (3, −1,8) to (5,7, −9).
5. Find the displacement vector from (−2,5,1) to (8, −1, −6).
6. Find the point 60% of the way from (3,2, −4) to (13, −8,1).
7. Find the point 35% of the way from (6,2,1) to (4, −1, −3).
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Section
IV.C.4 Cross
Product
The cross product is distinctly different from the dot product in that a cross product results in a vector,
a vector that is extremely special. The cross product, a⃗ × b⃗, is perpendicular to both a⃗ and b⃗. We say
that it’s normal (another word for perpendicular) to the plane that contains a⃗ and b⃗ when they are
placed tail-to-tail (a requirement for vector multiplication). In this section, we’ll simply learn how to
calculate the cross product.
Cross Product
The easiest way to calculate the cross product is to calculate the determinant of a 3 × 3 matrix by
hand. For a quick review, we can look back at II.D.3. Also, as a quick reminder, ı⃗, ȷ⃗, and k⃗ are unit
vectors in the positive x, y, and z directions, respectively. A vector 〈5,2,3〉 can be written as
5ı⃗ + 2ȷ⃗ + 3k⃗.
Example 1
Calculate the cross product 〈2,3,5〉 × 〈6,2,4〉.
ı⃗ ȷ⃗ k ⃗
2 3 5 = 3(4) − 2(5)ı⃗ − 2(4) − 6(5)ȷ⃗ + 2(2) − 6(3)k⃗ = 2ı⃗ + 22ȷ⃗ − 14k⃗ = 〈2,22, −14〉
6 2 4
Note: Remember we cross the row and column containing ı⃗ and calculate the 2 × 2 determinant not
crossed out. We do the same thing for ȷ⃗ and k⃗, making sure to negate the ȷ⃗ term.
Example 2
〈2, −5,1〉 × 〈−4,2, −3〉 =
ı⃗ ȷ⃗ k ⃗
2 −5 1 = −5(−3) − 2(1)ı⃗ − 2(−3)— 4(1)ȷ⃗ + 2(2) − (−4)(−5)k⃗
−4 2 −3
= 13ı⃗ + 10ȷ⃗ − 16k⃗ = 〈13,10, −16〉.
Now for some practice…
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IV.C.4 Independent Practice
Calculate the following cross products.
1. < 3,5,6 > × < 4,8,2 > =
2. < 5, −2, −1 > × < −1,4,2 > =
3. < 8,2, −5 > × < 4, −5, −3 > =
4. < −4,3,1 > × < 8,7, −2 > =
5. < 9,3,6 > × < 6, −3, −2 > =
6. < 5,2, −4 > × < −3,1,0 > =
7. < −1, −3, −6 > × < 2, −4,9 > =
8. < 5,2,4 > × < −3, −4, −10 > =
9. < 0,4, −2 > × < 3, −5, −1 > =
10. < −4,7,5 > × < 6,1, −2 > =
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Section
IV.C.5
Cross Product
Applications
There are two main applications for cross products we will take a gander at in this section: calculating
the area of a triangle, knowing only the three dimensional coordinates of its vertices, and determining
the equation for a two dimensional plane in three dimensional space.
Area of a Triangle
We have calculated the area of triangles before, and, truth be told, we could calculate the lengths of
the sides using the “distance formula” and then calculate the area with Hero’s formula… But there’s a
much easier way to go about it. If we can connect two vectors tail-to-tail that line up as two sides of
the triangle, then
Example 1
Area = a⃗ × b ⃗
Calculate the area of the triangle with vertices a(3,2,4), b(5,1,7), & c(6,2,1).
We need two vectors that line up tail-to-tail, so we’ll find displacement vectors ab ⃗ and ac ⃗.
ab ⃗ = 〈5 − 3,1 − 2,7 − 4〉 = 〈2, −1,3〉 and ac ⃗ = 〈6 − 3,2 − 2,1 − 4〉 = 〈3,0, −3〉
ab ⃗ × ac ⃗ = ı⃗ ȷ⃗ k ⃗
2 −1 3 = 〈−3,3,3〉
3 0 3
Area = |〈−3,3,3〉| = (−3) + 3 + 3 = √27 = 2.598.
Equation of a Plane
The equation of a plane in three dimensional space is Ax + By + Cz = D, where 〈A, B, C〉 is a vector
that is normal to the plane. If only we had some way to calculate a normal vector… Additionally,
three non-collinear points define any plane.
Example 2
Determine the equation of the plane that contains the points a(3, −2,1), b(2,5, −4), & c(−2,5,1).
ab ⃗ = 〈−1,7, −5〉 & ac ⃗ = 〈−5,7,0〉
ab ⃗ × ac ⃗ = ı⃗ ȷ⃗ k ⃗
−1 7 −5 = 〈35,25,28〉
−5 7 0
The equation of the plane is 35x + 25y + 28z = D, but how do we calculate D? Well (x,y,z) is any
point in the plane, and we have three to choose from. 35(3) + 25(−2) + 28(1) = 83.
Therefore, the equation of the plane is 35x + 25y + 28z = 83. We can verify this equation with either
of the two other points, and we’ll get 83 every time.
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IV.C.5 Independent Practice
Calculate the areas of the triangles with the given vertices.
1. (3,2,1), (5,4,7), & (6,2,4)
2. (5, −2,4), (7,4, −6), & (−3,8,6)
3. (7,3, −1), (−3, −7,2), & (7, −1,9)
4. (5,3,2), (−5,3, −6), & (5, −7, −1)
5. (−2, −4, −7), (6,2,1), & (6, −3, −4)
6. (5, −3,1), (−4, −3,5), & (2,8, −7)
Determine the equations for the planes that contain the following points.
7. (5,1,7), (6,4,1), & (4,2,8)
8. (3, −1, −6), (−2,4,7), & (1,4, −7)
9. (2,5,1), (5, −3,7), & (−2,5,9)
10. (5,3, −2), (−5, −2, −8), & (0, −4,6)
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Unit
V
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Unit
V Unit Overview
The Purpose of the Unit
The purpose of this unit is to introduce advanced counting methods for calculating sample sets and
probabilities. This information will help lead to any statistics courses we may take in college.
This History of Combinatorics
The earliest form of combinatorics can be traced back to India in 300 BCE in a written text called the
Bhagabati Sutra, which actually was an ancient “cookbook” of sorts, asking how many different
flavors one could experience choosing one, two, or three tastes from six total. Binomial coefficients
were discovered by Pingala as a byproduct of musical investigation, which would be expanded by
Bhaskara and Hemacandra over one thousand years later. Bhaskara would develop the idea of
“choice” and Hemacandra would use music to develop the Fibonacci sequence (which would
eventually be named for a European mathematician who brought the idea from India to Italy).
The ideas of combinatorics would be investigated by China, Egypt, and Greece but not to the same
extend they had in India. China would use it to establish how many arrangements there are in the I
Ching and on the board of a game of Go. The Middle East developed the idea of symmetric binomial
coefficients, their connection to binomial expansion, and Pascal’s triangle. Even though Pascal
wouldn’t be born for another 600 years, the triangle is named for him due to the extensive proofs he
conducted in relation to it, which goes to show how important mathematicians consider proof.
The ideas of combinatorics migrated to Europe (along with our modern day numbering scheme)
thanks to Fibonacci and Jordanus de Nemore. Pascal and Liebnitz (one of the discoverers of
calculus) are considered the founders of modern combinatorics, due to their work with Pascal’s
triangle and partitions. Euler would use the ideas to develop concepts in probability and graph
theory.
Back in the 17 th century, Blaise Pascal and Pierre de Fermat developed the first probability theory to
describe likelihoods of outcomes in a game of chance. De Moivre and Bernoulli would go on to
developed probability as a sound field of mathematics. As a result, binomial probability is known by
statisticians as Bernoulli trials.
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Chapter
V.A
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Chapter
V.A Chapter
Overview
‣ V.A.1 – Factorial Arithmetic
This section introduces the factorial, using them to simplify advanced multiplication and division
problems.
‣ V.A.2 – Counting Choices
This section explains the concept of choice and how we can use it to calculate totals without
counting each individual possible outcome.
‣ V.A.3 – Rearrangements
This section calculates all of the possible rearrangements of a sequence of elements including
and excluding repetition.
‣ V.A.4 – Permutations
This section defines permutations in terms of counting choices and rearrangements from the
previous two sections in terms of factorials.
‣ V.A.5 – Combinations
This section utilizes the previous four sections to understand the concept of combinations in
preparation for probability in the next chapter.
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Section
V.A.1 Factorial
Arithmetic
In III.E, we dealt with sequences and series of values, where series were the sums of sequences. A
factorial is a special form of a product of sequences, and we use an exclamation point, “!”, to denote a
factorial.
Example 1
Calculate 5!
7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 5040.
n! = n(n − 1)(n − 2) … (3)(2)(1)
Factorials follow all rules of fractions, where we can cancel out common factors from the numerator
and denominator.
Example 2
Rewrite
!
∙∙
as a single factorial.
!
= ∙∙∙!
= 75! .
∙∙ ∙∙
Note: As we can see in the previous example we can expand a factorial as much as or as little as we
wish.
Example 3
Calculate !
! .
!
= ∙∙!
= 95 ∙ 94 = 8930.
! !
Example 4
Write 20 ∙ 19 ∙ 18 in terms of factorials.
As we can see 20 ∙ 19 ∙ 18 are the first terms in 20!, but we’re missing the other 17 terms. Therefore
20 ∙ 19 ∙ 18 = !
! .
Now, we’re all set for the rest of the chapter. See? Factorials are exciting!
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V.A.1 Independent Practice
Expand the following factorials and calculate the numerical value.
1. 1! =
4. 4! =
2. 2! =
5. 5! =
3. 3! =
6. 6! =
Rewrite the following as a single factorial.
7. 5! ∙ 6 ∙ 7 =
10. !∙∙
=
8.
!
∙∙ =
11.
!
=
∙∙∙
9. ∙∙
=
12. 120 ∙ 56 =
Simplify the following and calculate the numerical value.
13. !
! =
16. !
! =
14. !
! =
17. !∙!
!
=
15. !
!
=
18. !
(!) =
Rewrite the following in terms of factorials.
19. 95 ∙ 96 ∙ 97 =
∙∙∙∙ =
20. 36 ∙ 37 =
21. ∙∙
22.
23. 121
24.
∙
∙
120 ∙ 119 ∙ 118 =
∙∙ =
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Section
V.A.2
Counting Choices
To determine the total number of possible outcomes, we must understand the concept of choice.
Example 1
A license plate used to have six characters. Originally, the first three were letters and the last three
were numbers. How many different license plate “numbers” are possible?
We have six “spaces” to fill. The first space has 26 choices, as do the second and third spaces. The
fourth, fifth, and sixth spaces have 10 choices (0-9) each. Therefore, there are
26 ∙ 26 ∙ 26 ∙ 10 ∙ 10 ∙ 10 = 26 ∙ 10 = 17,576,000 possible license plate numbers.
Example 2
A license plate now has seven characters, and numbers and letters can appear in any order.
However the letters I and O are not allowed due to their resemblance to 1 and 0. Additionally, in
certain states, they do not allow repetition of characters. How many plate “numbers” are possible?
There are 26 + 10 − 2 = 34 possible characters that we can use. The first space, then, has 34
choices. Due to the fact that we cannot repeat elements, once the first element has been chosen,
there are only 33 to choose from for the second space, and so on. Therefore, there are
34 ∙ 33 ∙ 32 ∙ 31 ∙ 30 ∙ 29 ∙ 28 = !
! = 2.711 × 10 possible license plate numbers, quite a few more
than what we found in Example 1.
Example 3
A new pizza place offers four types of meats, seven types of vegetables, and three types of cheese,
and two types of crusts. How many different pizzas can be created if we have one choice from each
category? Keep in mind, we can have a meatless pizza or a pizza with no vegetables. However,
pizzas must have crust and cheese. Otherwise it’s no pizza we want.
There are four “spaces” in this problem. The first is 5 (4 meats and 1 no-meat), the second is 8 (7
veggies and 1 no-veggie), the third has 3, and the last has 2. Therefore, there are 5 ∙ 8 ∙ 3 ∙ 2 = 240
possible pizzas!
Although, it won’t come up in this section, there is a distinct difference between the words “and” and
“or”. In a word problem concerning counting, “and” indicates multiplication, and “or” indicates
addition. All of the above problems and problems in the practice set are “and” problems, hence we’ve
only multiplied. We’ll talk about “or” in the next chapter when we deal with probability.
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V.A.2 Independent Practice
Answer the following questions.
1. Debbie wants to open a sandwich shop. She will offer 5 different meats and 3 different
cheeses and 7 different breads. How many sandwiches can she offer her customers?
2. The local ice cream shop offers 4 different cones, 20 different flavors of ice cream, and 8
different “mix-ins”, of which a customer can only choose one of each. How many ice cream
combinations can one order?
3. A combination lock has 50 numbers on it. If a combination contains three numbers, how many
combinations are possible?
4. In the smaller states, a license plate has three letters followed by three numbers. The letters O
and I are not used because of their resemblance to 0 and 1. How many license plates can be
issued?
5. Many alarm codes are made up of four digits ranging from 0 to 9. How many codes are
possible?
6. In order to make the code more difficult to crack, George decides to not repeat numbers in his
code. Out of how many codes does George have to choose one? Does the elimination of
repetition increase or decrease his choices?
7. There are 15 six-year-olds at a birthday party. If there are 4 kinds of party hats to choose
from, how many possible ways are there for party hats to be dispensed to the kids?
8. There are 6 chairs, and the 15 children are playing musical chairs. How many ways can the
chairs be filled at the end of the song?
9. How many possible combinations of initials can a person have?
10. If Dave tosses 3 different colored 6-sided dice, how many results are possible?
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Section
V.A.3 Rearrangements
In this section we will examine all rearrangements of a set of finite data. In simplistic terms, it’s like
finding the number of different anagrams one can make from a given word. In this unit, the term
“words” indicates rearrangements of a given set of letters. They do not have to be pronounceable.
Example 1
How many ways are there to rearrange the letters in FERMAT?
There are six spaces, six choices for the first, five for the second, and so on. Therefore, there are
6! = 720 possible rearrangements.
This leads to the fact that if a sequence has n unique terms, there are n! ways to rearrange the
sequence. This is all well and good, but as we know, many words in the English language have
repeated letters in them. What do we do about those?
Example 2
How many ways are there to rearrange the letters in BHASKARA?
If all 8 letters in this problem were unique, there would be 8! = 40,320 different rearrangements.
However, there are 3 A’s in the word. If each A were unique, there would be 3! = 6 rearrangements
of them, which means each 8-letter rearrangement is actually being counted 6 times each. Therefore
there are !
=
= 6,720 different rearrangements of BHASKARA.
!
Example 3
How many ways are there to rearrange the letters in FIBONACCI?
In this 9-letter word there are 2 I’s and 2 C’s. Therefore, there are !
Example 4
How many ways are there to arrange 4 guests around a circular table?
!!
= 90,720 rearrangements.
Firstly, looking at the tables at right, because there is no head of a circular
D 1 B C 2 A B 3
table, tables 1 and 2 have the same arrangement. They’re just rotations.
C
B
C
Table 3 is a completely different arrangement. Since all four people will sit at the table, we can
choose one to be our point of reference. If we choose A as our point of reference, then we have 3
choices for the next clockwise position, 2 for the next, and 1 for the last. So for four people sitting
around a round table there are only 3! = 6 different arrangements.
Note: A key ring problem is a tricky form of the round-table problem. With a key ring, arrangements 1
and 3 above would be considered the same since the key ring can flip over. In a key ring problem,
we would have to divide by 2 as our last step.
A
D
A
D
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V.A.3 Independent Practice
Answer the following questions.
1. How many ways are there to rearrange the letters in WACO?
2. How many ways are there to rearrange the letters in AUSTIN?
3. How many ways are there to rearrange the letters in HOUSTON?
4. How many ways are there to rearrange the letters in DALLAS?
5. How many ways are there to rearrange the letters in SAN ANTONIO?
6. If 10 people are running in a race, in how many ways can the 10 people finish?
7. In how many ways can 8 people sit around a table?
8. In how many ways can 4 men and 4 women sit around a table such that no man sits next to
another man and no woman sits next to another woman?
9. In how many ways can 10 keys be arranged on a key ring?
10. If a coach needs to put together a 9-person team, randomly assigning positions, how many
teams could the coach create?
11. If you have 16 different colored billiard balls, in how many ways can they be drawn randomly
from a bag?
12. There are 9 books on a shelf. If two of the books must be next to one another no matter what,
in how many ways can the books be arranged?
13. In how many ways can the 9 books be arranged if 3 of the books must be set together?
14. Doug’s dog, Wanda, has 8 puppies. There is a white puppy, a brown puppy, and 6 black
puppies. If the six black puppies are identical, in how many ways can he arrange the puppies
in a row for a family picture?
15. Dorian finally got a girl’s phone number, but he tried to impress her by memorizing it. The next
day all he can remember is that there are 2 threes, 3 eights, and 2 fives. How many different
numbers will he have to try to guarantee he calls the right person?
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Section
V.A.4 Permutations
A permutation is the same thing as an arrangement, however, traditionally, we relegate the term to
indicate no repeated elements in the population and a sample set of any size less than or equal to
that of the overall population. What does that mean? We’re not restricted to having to use every
element given to us, and we won’t have repetitions.
Example 1
How many four-letter words can be made from the letters in FERMAT?
Since we’re creating four-letter words from six letters, this is like having four blank spaces to fill. We
have 6 choices for space 1, 5 choices for space 2, 4 choices for space 3, and 3 choices for space 2.
Therefore, there are 6 ∙ 5 ∙ 4 ∙ 3 = !
= 360 different possible words.
Notation
!
Traditionally, the number of permutations of size r from a set of size n is written as P , and as we
can see in Example 1,
Example 2
P = !
()!
Fifty people are running in a marathon. There are prizes for the top 10 places. In how many ways
can there be a top 10?
Essentially, this is asking us to calculate
P =
!
= !
= 50 ∙ 49 ∙ 48 ∙ … ∙ 42 ∙ 41 = 3.728 × ()! ! 10 .
Example 3
How many ways can Bob arrange 5 rubber ducks in a row given 10 rubber ducks to choose from.
There are P = !
!
= 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 = 30,240 arrangements of rubber ducks.
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V.A.4 Independent Practice
Answer the following questions.
1. 8 horses are running a race. In how many ways could there be a different first, second, and
third place?
2. 10 people are running for four offices: President, Vice President, Secretary, and Historian. If
any person can earn any position, how many possible ways can the offices be assigned?
3. How many four letter ‘words’ can be created from the letters in MONKEY without repetition?
4. How many three letter ‘words’ can be created from the letters in BACHELOR without
repetition?
5. If there are 8 available seats on a bus, and 15 people get on, in how many ways can the seats
be filled?
6. Bill has a bag filled with 10 different-colored markers. If you need four markers, in how many
ways can the markers be pulled out one at a time?
7. If there is room on a shelf for only 5 out of 12 different rubber ducks, how many ways can
rubber ducks be arranged on a shelf?
8. Out of a deck of 52 cards, in how many ways can you draw 5 cards one at a time?
9. If a license plate has 3 letters and 3 numbers and no repetitions are allowed, how many
license plates can be produced (Reminder: no O’s or I’s)?
10. In a video game, how many ways are there for four people to choose a character from a roster
of 35 if no character can be chosen more than once?
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Section
V.A.5 Combinations
Combinations are exactly like permutations except for one thing: THE ORDER DOES NOT MATTER.
This means the combination ABC is the same as CBA, ACB, CAB, BCA, and BAC. These six
permutations are the same combination. So, turn to turn an P into a C , we need to divide
.
Keep in mind, there are six rearrangements of ABC because 3! = 6.
Example 1
How many ways can 4 letters be chosen from ABCDEFG?
There are P = !
= 7 ∙ 6 ∙ 5 ∙ 4 = 840 permutations. We must divide by the 4! different arrangements
!
of each possible combination of four elements. So, there are
= 35 combinations.
Notation
There are two notations that mean combination: C and n . Both are read as “n choose r”. The
r
formula for combinations is
Example 2
C = n r =
!
!
= !
()!!
How many ways are there to choose an 8 person dodge ball team from a set of 16 people?
Since the order doesn’t matter, there are 16
8 = !
= 12,870 different dodge ball teams possible.
!!
Example 3
If there are 10 men and 8 women in a group, how many ways are there to select 4 men and 3
women?
10
4 8 3 = !
∙ !
= 11,760 ways.
!! !!
Now, in the previous problem we can see that in 10
4 = !
, the two numbers in the bottom sum to
!!
the top number. Therefore, 10
4 = 10 , and it’s this instance of symmetry that Pascal’s Triangle is
6
based upon.
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Pascal’s Triangle
Here are the first 6 rows of Pascal’s Triangle:
n = 0: 1
n = 1: 1 1
n = 2: 1 2 1
n = 3: 1 3 3 1
n = 4: 1 4 6 4 1
n = 5: 1 5 10 10 5 1
Each term in Pascal’s Triangle is the sum of the two terms above it, in a recursive way. So the n = 6
row would be 1 6 15 20 15 6 1. The first term in each row is r = 0, and the last is r = n. Therefore,
4 would be on the n = 4 row, and r = 3 would be the fourth element in the row, 4. Also, Pascal’s
3
triangle reveals certain nifty facts: n 0 = n n = 1 and n 1 = n
n − 1 = n.
Calculator Interlude
Factorials, Permutations, and Combinations can be performed in the TI-84 calculator. If we go to
MATH and scroll over to the PRB menu we see “2: nPr”, “3:nCr”, and “4:!”. To use nPr and nCr, we
input the n value into the home screen, choose nPr or nCr, then input r, and press ENTER. The
factorial sign is used normally.
Now that we’re approaching more difficult problems, it’s vital we learn alternate ways to calculate
combinations, as they are more prolific than permutations.
Poker Problems
One of the most interesting and complicated forums for combinations is poker hands. Traditionally,
mathematicians concentrate on 5 card hands, specifically 5-card stud since no cards can be traded
back. In order to master poker problems, we must understand a standard deck of playing cards: 52
cards made up of thirteen different values (A, 2-10, J, Q, K) in four suits (red hearts, red diamonds,
black spades, and black clubs). We usually ignore the two jokers.
Example 4
In how many ways can a three-of-a-kind be dealt in a game of 5-card stud?
A three-of-a-kind is a hand with three cards of the same value and two random cards that don’t match
the three-of-a-kind’s value nor each other’s. So first we need to choose the value of the three of a
kind 13
, then we have to choose
1 , then we have to choose the three suits of the three of a kind 4 3
the last two values from the 12 remaining values 12 , and lastly we choose the suits for each of the
2
singletons 4 1 . 13
1 4 3 12 2 4 1 4 = 13 ∙ 4 ∙ 66 ∙ 4 ∙ 4 = 54,912 possible three-of-a-kinds.
1
Note: The value of the three-of-a-kind must be chosen separately from the other two. The two
singletons must be chosen at the same time because they’re both one-of-a-kinds. If we were asked
for two pairs, we would choose both values at the same time 13 . If we consider the number of
2
cards of the same value as the degree, values of the same degree will be chosen together.
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V.A.5 Independent Practice
Answer the following questions.
1. In how many ways can a school board of 6 people be chosen from a pool of 18 candidates?
2. How many ways can 4 different-colored markers be pulled from a bag of 10 all at the same
time?
3. If 100 people enter a drawing, and 8 people can all win the same prize, in how many ways can
those 8 people be chosen?
4. How many 5-card poker hands are in a deck of 52 cards?
5. In how many ways can 9 books out of 30 be chosen for a book shelf?
6. In how many ways a coin be flipped 20 times and come up heads 12 of those times?
7. How many ways can 3 letters be chosen from the word GARBLED?
8. A certain pizza parlor is offering pizzas with 2 meats and 3 veggies. If it has 6 meats and 10
veggies to choose from, how many pizzas can be created?
9. How many ways can a full house be drawn in a game of 5 card stud?
10. How many ways can two pair be chosen in a game of 5 card stud?
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Chapter
V.B
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Chapter
V.B Chapter
Overview
‣ V.B.1 – Probability Conceptually
This section will introduce notation for probability and multiple statistical terms. It will also
describe in detail the conceptual nature of probability.
‣ V.B.2 – Probability Graphically
This section will demonstrate graphical representations of probability, as well as conditional
probability.
‣ V.B.3 – Probability of Permutations & Combinations
This section will apply the information from V.A to calculate advanced probabilities.
‣ V.B.4 – Binomial Probability
This section will utilize Pascal’s Triangle to explain probability of success and failure.
‣ V.B.5 – Mathematical Expectation
This section will describe mathematical expectation in terms of games of chance, something that
the study of probability is based on.
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Section
V.B.1 Probability
Conceptually
Now that we understand basic counting principals, it’s time we put them to use. In this section, we
will introduce the notation and concepts of probability we will use throughout the chapter.
What is Probability?
Probability is the likelihood of an event occurring. To calculate it, we divide the number of ways to
achieve a specific event by the number of total possible events. For instance, the probability of
getting a royal flush in poker can be calculated as
P(Royal Flush) =
=
= 1.539 × 10 = 1.539 × 10 % because there are only 4 possible
royal flushes in a game of poker and there are 52 total possible random 5-card hands. Generally,
5
Intersections
P(specific outcome) =
An intersection is our friendly neighborhood “and”, which, if we remember, implies multiplication.
P(A ∩ B) is read “the probability of A and B” and means “the probability that both events A and B will
occur. To calculate the probability of an intersection:
P(A ∩ B) = P(A)P(B|A)
This says the probability of A and B is equal to the probability of A times the probability of B given that
A has already occurred.
Example 1
Given a bag containing 20 red poker chips, 15 blue poker chips, and 10 white poker chips, what is the
probability of drawing two red poker chips and a white poker chip without replacement?
If we let R = drawing a red poker chip, and W = drawing a white poker chip, then the problem is
asking P(R ∩ R ∩ W) = P(R)P(R|R)PW(R ∩ R). There are 45 chips total, so P(R) =
. Since
we’re not replacing chips after we’ve drawn them, P(R|R) =
: one less red chip out and one less
chip total. Finally, PW(R ∩ R) =
as there are still 10 white chips, and now only 43 chips total
(two have been drawn). P(R ∩ R ∩ W) =
Example 2
= .045 = 4.463%.
Given the same situation in Example 1, what is the probability of drawing two red poker chips and a
white poker chip with replacement?
Since the chips are replaced after being drawn, P(R) =
P(R ∩ R ∩ W) =
, P(R|R) = , PW(R ∩ R) = . Thus
= .0439 = 4.390%, which is very close to the answer in Example 1.
In Example 2, the three situations are considered independent because the outcomes have no
influence on each other. If two events A and B are independent, then P(A ∩ B) = P(A)P(B).
252
Cogswell – PCM

Unions
A union is an “or” operation, which implies an addition operation. P(A ∪ B) is read “the probability of
A or B” and means “the probability of A occurring, B occurring, or both”. In both statistics and logic
the term, “or”, is not exclusive. It’s actually an “and/or”.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
The significance of the subtracted intersection, P(A ∩ B), comes about because any element
contained within the intersection will be counted once in P(A) and again in P(B). We can’t allow any
element to be counted twice, so we have to subtract one of the repeated instances. This fact will be
made much clearer in the next section.
Example 3
Every student in a small school is required to be in band or choir. If there are 50 students in band, 40
students in choir, and 20 students in both, what is the probability of choosing a band student at
random?
Let B = Students in Band, and C = Students in Choir. To calculate the probability of choosing one of
the 50 band students, we must know how many students there are total. This works exactly like
probability: n(B ∪ C) = n(B) + n(C) − n(B ∩ C) = 50 + 40 − 20 = 70. Therefore P(B) =
.
Example 4
There are 30 students on the football team, 25 students on the basketball, and 15 students on the
swim team. No students are allowed to be in more than one sport. What is the probability of a
student chosen at random being a member of the basketball team or swim team?
Let B = basketball player and S = swimmer. Then P(B) =
, P(S) = , P(B ∩ S) = 0. Then
P(B ∪ S) =
+
=
= .
In Example 2, P(B ∩ S) = 0, because no student can be in more than one sport. In this instance
Basketball and Swimming are considered mutually exclusive, since there is no overlap. In general,
if P(A ∩ B) = 0, then A and B are mutually exclusive, and P(A ∪ B) = P(A) + P(B).
Negations
The tilde symbol, “~”, translates to “not”, and P(~A) = 1 − P(A).
Example 5
If there are 30 students in a club, and 5 them are the leaders, what is the probability of choosing a
non-leader at random?
P(~L) = 1 − P(L) = 1 − =
= .
253
Cogswell – PCM

Important Facts
The probability of a certainty is 1. For instance if B = Boy and G = Girl, then the probability of
choosing a boy or girl at random is P(B ∪ G) = 1 or 100%. This is true because B ∪ G contains all
members of the set.
The probability of an impossibility is 0. For instance if B = Boy and G = Girl, then the probability of
choosing a person at random who is both a boy and a girl is P(B ∩ G) = 0 because there are no
people who are both. We can argue these two points, of course, due to the nature of
hermaphrodites, but we’ll save that discussion for biology, anatomy, or genetics.
Negations are actually quite powerful devices if we truly understand them. Many people mistakenly
believe that “all” is the negation of “none”. This is not the case. In actually, the negation of “none” is
“at least one”. Let’s say, for instance, that we flip 5 coins with an outcome of either heads or tails.
P(H = n) is the probability of flipping n heads, where n is any number from 0 to 5 in our example. If
we’re asked to find P(H ≥ 1), we could either find P(1 ∪ 2 ∪ 3 ∪ 4 ∪ 5) = P(1) + P(2) + P(3) + P(4) +
P(5), which may take a while, or we can understand that P(H ≥ 1) = 1 − P(0), which would be much
easier to find.
Generally, ~(A ∪ B) = ~A ∩ ~B, and ~(A ∩ B) = ~A ∪ ~B.
Also, we have shown P(A ∪ B) = P(A) + P(B) − P(A ∩ B), but for three elements
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C)
which will be explained in detail in the next section.
254
Cogswell – PCM

V.B.1 Independent Practice
Answer the following questions.
1. In a study of 1200 people chosen at random, 832 liked dogs more than cats (d), 256 liked cats
more than dogs (c), and the remaining subjects liked neither equally (n). The same 1200
people were asked if they would be willing to spend more than $30 on a special collar. 532
said ‘Yes’ (y), and everyone else said ‘No’ (~y).
a. P(d) =
g. P(d ∩ y) =
b. P(c) =
h. P(c ∩ ~y) =
c. P(n) =
i. P(d ∪ c ∪ n) =
d. P(y) =
j. P(c ∪ d) =
e. P(~y) =
k. P(d ∪ ~y) =
f. P(d ∩ c) =
l. P(c ∪ y) =
2. In a study of 250 households, 142 owned 3 or more televisions (t), and the rest owned less
than 3 televisions (~t). In the same 250 households, 112 watched comedies most often (c), 56
watched dramas most often (d), and the rest preferred nature shows (n).
a. P(t) =
h. P(~t ∩ d) =
b. P(~t) =
i. P(t ∩ n) =
c. P(c) =
j. P(c ∩ d) =
d. P(d) =
k. P(c ∪ d) =
e. P(n) =
l. P(d ∪ n) =
f. P(t ∩ ~t) =
m. P(t ∪ c) =
g. P(t ∩ c) =
n. P(~t ∪ n) =
255
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Section
V.B.2 Probability
Graphically
In this section, we will represent the information gleaned in the last section graphically.
Venn Diagrams
Let’s start with 2-circle venn diagrams.
Example 1
Given the venn diagram at right, calculate P(A), P(B), P(~A), P(~B),
P(A ∩ B), P(A ∪ B), P(~A ∪ ~B), P(~A ∩ ~B).
25
A
21 15 13
B
Firstly, let’s investigate the venn diagram. 21 represents A ∩ ~B (only A), 15 represents A ∩ B, 13
represents B ∩ ~A (only B), and 25 represents ~A ∩ ~B. The combined values within the circles
represent A ∪ B. Additionally, the total number of elements is 25 + 21 + 15 + 13 = 74.
P(A) =
P(B) =
P(~A) =
P(~B) =
=
,
=
,
=
= 1 − P(A),
= 1 − P(B), P(A ∩ B) =
=
P(A ∪ B) =
=
P(~A ∪ ~B) =
P(~A ∩ ~B) =
,
= P(A) + P(B) − P(A ∩ B),
=
= 1 − P(A ∪ B).
= 1 − P(A ∩ B), and
Hopefully, by now we understand the concept of negation. Now let’s move on to 3-circle venn
diagrams.
Example 2
Given the venn diagram at right, calculate P(A), P(B), P(C), P(A ∩ B), P(B ∩ C),
P(A ∩ C), P(A ∩ B ∩ C), and P(A ∪ B ∪ C).
There are 21 + 15 + 13 + 14 + 5 + 32 + 25 = 125 elements.
P(A) =
=
,
P(B) = ()
P(C) =
P(A ∩ B) =
P(B ∩ C) =
,
P(A ∩ C) =
P(A ∩ B ∩ C) =
=
,
=
,
=
,
=
,
, and
21 + 15 + 13 + 14 + 5 + 32
P(A ∪ B ∪ C) = = 100
125
125
P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C) =
+
+
−
−
−
+
=
,
1 − P~(A ∪ B ∪ C) = 1 −
=
.
25
A
21 15 13
5
14 0
32
C
B
256
Cogswell – PCM

Tree Diagrams and Conditional Probability
Conditional probability has already been discussed when we introduced P(A|B), which is the
probability of A given that B has already occurred. B occurring is the condition.
Example 3
Many athletes are superstitious. For instance, Doug, a football player, thinks that if his team wins the
coin toss, he will be more likely to score a touchdown in the game. This thought actually creates a
physical outcome. Based on his stats, if the team wins the coin toss, Doug has a 70% of scoring a
touchdown, and if the team loses the coin toss, Doug has only a 35% chance of scoring a touchdown.
What is the probability that (a) the team will win the coin toss and Doug will score a touchdown, (b)
the team winning the coin toss and Doug not scoring a touchdown, (c) the team not winning the coin
toss and Doug not scoring a touchdown, and (d) the team losing the coin toss and Doug scoring a
touchdown?
We can easily express this as a tree diagram:
70%
touchdown
1/2
win
30%
no touchdown
coin toss
35%
touchdown
1/2
lose
65%
no touchdown
If W = win, then ~W = lose, and if T = touchdown, then ~T = no touchdown.
(a) P(W ∩ T) = P(W)P(T|W) = (. 5)(. 7) = .35 = 35%
(b) P(W ∩ ~T) = P(W)P(~T|W) = (. 5)(. 3) = .15 = 15%
(c) P(~W ∩ ~T) = P(~W)P(~T|~W) = (. 5)(. 65) = .325 = 32.5%
(d) P(~W ∩ T) = P(~W)P(T|~W) = (. 5)(. 35) = .175 = 17.5%
All we have to do is multiply along the branches.
Note: If we add all probabilities together in Example 1, we get 100%. If we don’t, we’ve done
something wrong.
257
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V.B.2 Independent Practice
Describe the meaning of each probability and calculate each percentage.
1. The following Venn Diagram describes the result of a study of students who are involved in
B
athletics (A), band (B), and Choir (C).
A
a. P(A) =
i. P(~A ∪ C) =
b. P(B) =
j. P(A ∩ B ∩ C) =
23 31 30
18
14 20
25
19
C
c. P(C) =
d. P(~A) =
e. P(~B) =
f. P(~C) =
g. P(A ∪ B ∪ C) =
h. P(A ∪ ~B) =
k. P(A ∩ ~C) =
l. P(~B ∩ C) =
m. P(~A ∩ ~B ∩ ~C) =
n. P(A ∩ B) ∪ C =
o. P(A ∩ ~B) ∩ C =
p. P(A ∪ B) ∩ ~C =
2. The following is a conditional probability tree diagram that describes the results of a study on ice
cream flavors. Men (A) and Women (~A) were asked if the preferred vanilla (B), chocolate (C), or
36% B
strawberry (D).
a. P(A) =
b. P(~A) =
j. P(A ∩ B) =
k. P(A ∩ C) =
65%
35%
A
~A
52%
12%
C
D
43% B
27%
C
30%
D
c. P(A ∪ ~A) =
d. P(B|A) =
e. P(C|A) =
f. P(D|A) =
g. P(B|~A) =
h. P(C|~A) =
i. P(D|~A) =
l. P(A ∩ D) =
m. P(~A ∩ B) =
n. P(~A ∩ C) =
o. P(~A ∩ D) =
p. P(B) =
q. P(C) =
r. P(D) =
258
Cogswell – PCM

Section
V.B.3
Probability Using
Permutations & Combinations
In this section, we’re going to delve back into the permutations and combinations we learned back in
V.A, and use them to calculate probability.
Example 1
A blind person wants to organize a series of 6 identically-shaped books on a shelf. What is the
probability that he will put the first book first and last book last?
This is a rearrangement problem, and because he’s blind, there are 6! = 720 different possible
arrangements. We want to know how many ways there are for the first book to be first and the last
book to be last. If we place those two accordingly, there’s still four blank spaces in between to fill with
4! = 24 different arrangements. Therefore the probability is !
=
= .
! ∙
Example 2
Three men and three women will sit at a round table. What is the probability they will sit boy-girl?
We know from V.A.3, that if all six people sat at the table randomly, there would be 5! = 120 different
seating arrangements. To determine how many of these are boy-girl, we select one person to be our
point of reference. Assuming we’ve chosen a man, then the seat next to him has to be one of the 3
women, the next seat must be one of the remaining 2 men, the next seat will be one of the 2
remaining women, and the last two seats must be the remaining man and woman. Therefore there
are 3(2)(2)(1)(1) = 12 boy-girl seating arrangements. The probability is
= .1 = 10%.
Example 3
259
=
11 slips of paper numbered 1 to 11 are in a hat. If Roberta draws 3 slips, what is the probability 1 will
be odd and 2 will be even?
Firstly, there are 11 = 165 possible ways for Roberta to draw 3 slips of paper, randomly. Since
3
there are 6 odd numbers and 5 even numbers, there are 6 = 6 ways to draw and odd number and
1
5 ()
= 10 ways to draw two even numbers. Since this is an “and” problem, the probability is
2 =
.
Example 4
What is the probability of being dealt a flush in a game of 5-card stud?
A flush is a hand where all 5 cards share the same suit. There are 4 = 4 ways to choose the suit,
1
13 = 1287 ways to choose the 5 cards in that suit, and 52 = 2598960 possible hands. The
5 5
probability is ()
= 0.199%! That’s highly unlikely.

V.B.3 Independent Practice
Answer the following questions.
1. What is the probability of rolling a total of 7 when rolling two six-sided dice?
2. 10 horses are running in a race. What is the probability that Runsalot will come in first?
3. If you buy an 8-pack of markers, what is the probability that red will be first and green will be
last in the package?
4. If you have a bag with 6 different colored billiard balls, what is the probability you will draw out
the red ball first, orange ball third, and green ball fourth without replacing each ball after you
draw it. What is the probability of the same thing happening if you replace the ball back in the
bag after drawing it?
5. If you randomly scramble the letters in BACHELOR, what is the probability you will have a
‘word’ where the vowels stay in the same place?
6. In a door prize drawing with 40 entries, 3 names are to be drawn. What’s the probability
Georgette’s name will be drawn?
7. A combination lock has 50 numbers on it. If the combination has three different numbers, what
is the probability of getting none of the three numbers correct? What is the probability of
getting at least one number correct? What is the probability of getting all three numbers
correct?
8. If 12 men and 16 women are running for a committee of 10 people, what is the probability that
the committee will be made up of 5 men and 5 women?
9. Teddy has a penny, a nickel, a dime, a quarter, a half dollar, and a silver dollar. If he selected
two from his pocket, what is the probability of not choosing the dime?
10. If Harold draws 3 cards from a standard deck of 52 cards, what is the probability he will draw
two 5’s and a face card?
11. What is the probability of getting dealt a royal flush in a game of 5 card stud?
12. What is the probability of getting dealt a full house in a game of 5 card stud?
260
Cogswell – PCM

Section
V.B.4
Binomial Probability
In this section we’re going to look at the binomial theorem, its relationship with Pascal’s Triangle,
and what it all has to do with probability. Here’s a quick glance back at Pascal’s Triangle. Take a
good look at it we can see the connection in a couple minutes.
We have seen in I.C.1, that (a + b) = a + 2ab + b . Furthermore,
(a + b) = a + 3a b + 3ab + b and (a + b) = a + 4a b + 6a b + 4ab + b .
Made the connection yet? If not, take a gander at the coefficients. They match the row in Pascal’s
Triangle corresponding to the exponent. Additionally, the exponents of each term in the expansion
add up to that same exponent as well, with a’s exponent decreasing each time and b’s increasing to
compensate. We also know from V.A.5, that the values in Pascal’s Triangle correspond to
combinations. Thus, (a + b) = 3 0 a + 3 1 a b + 3 2 ab + 3 3 b , where the n value matches the
exponent of the binomial, and the r value matches the exponent of b in each term.
This is essentially the binomial theorem, and the actual simplification is called binomial expansion,
since we’re applying exponents to binomial expressions. Now, what in the world does all of this have
to do with Binomial Probability?
Binomial Probability
Binomial probability concerns instances with only two possible outcomes. In statistics, we refer to
these as success and failure, however, doing so sometimes result in awkward phrases. For instance,
if we’re trying to determine the probability of someone having cancer, according to statisticians,
getting cancer is a success. See? Awkward…
Let’s instead think of binomial probability as a coin flip with a severely bent coin. Because the coin is
bent, the weight is not evenly distributed. After much coin flipping, we have come to the conclusion
that the coin is 65% likely to land heads-up, so it is 35% (100% − 65%) likely to land tails-up. This is
a binomial probability problem just waiting to happen because there are only two possible outcomes.
Example 1
n = 0: 1
n = 1: 1 1
n = 2: 1 2 1
n = 3: 1 3 3 1
n = 4: 1 4 6 4 1
n = 5: 1 5 10 10 5 1
If we flip the coin described above 8 times, what is the likelihood it will land heads-up 4 times?
If we let H = heads-up and T = tails-up. This problem is asking for
P(H ∩ H ∩ H ∩ H ∩ T ∩ T ∩ T ∩ T). Because one coin flip has no effect on any other coin flip, the flips
are independent. So, we have to find P(H)P(H)P(H)P(H)P(T)P(T)P(T)P(T) = P(H) P(T) =
(. 65) (. 35) = .00267 = 2.67%, however we’ve actually calculated the probability of getting four
heads and then four tails, which is not what the problem is asking. Order doesn’t matter in this
problem. If we picture the 8 flips as blank spaces, we know 4 of them will be heads and 4 will be tails.
If we choose the four spaces to be heads, then the remaining four have to be filled with tails.
Therefore, there are 8 = 70 ways to get 4 heads and 4 tails, and the probability of this occurring is
4
. 00267(70) = .188 = 18.751%.
261
Cogswell – PCM

In Example 1, we found that the probability of getting 4 heads in 8 flips of our special coin was
8 4 P(H) P(T) , which should look awfully familiar. This is the r = 4 term of the n = 8 degree
expansion of the binomial P(H) + P(T). In other words we expand P(H) + P(T) and pick out the
term that has 4 H’s.
Example 2
If we flip our bent coin 7 times, what is the probability of getting 3 Heads?
Expanding P(H) + P(T) , we want the 7 4 P(H) P(T) term, which is 35(. 65) (. 35) = .144 =
14.424%. This small probability is understandable since the probability of a head is almost twice as
likely as that of a tail.
If, in Example 2, we wanted to find the probability of each different outcome, the probabilities would
always add up to 1 or 100%.
Example 3
If we flip our bent coin 5 times, what is the probability of getting 1 or 2 heads?
1 head and 2 heads are mutually exclusive events, so we will have to calculate them separately and
add the probabilities together. 5 4 (. 65) (. 35) = .0488 and 5 3 (. 65) (. 35) = .181. So
P(1 ∪ 2) = .0488 + .181 = .230 = 22.992%.
Example 4
If we flip the bent coin 10 times, what is the probability of getting at least 1 tail?
We could calculate all of options from 1 tail to 10 tails and add them together, but that would take a
considerable amount of time. However, we know that at least 1 tail is the negation of 0 tails. So if we
calculate the probability of no tails, then we can subtract the probability from 1.
P(T ≥ 1) = 1 − P(T = 0) = 1 − 10
0 (. 65) (. 35) = .987 = 98.654%. This may seem large, until we
realize that at a 35% tail rate, statistically, we should get a tail approximately 3 out of the 10 times,
which is greater than 1 time. Therefore, it’s almost a guarantee we will get at least one.
Note: 5 4 = 5 because 5 flips choose 4 heads is the same as 5 flips choose 1 tail.
1
262
Cogswell – PCM

V.B.4 Independent Practice
Answer the following questions.
1. Kathryn is genetically disposed to having daughters. If a doctor estimates that she has a 65%
chance of having a daughter and she plans on having 6 children, what is the probability
a. She will have all six daughters?
b. She will have three daughters?
c. She will have no daughters?
d. She will have at least one daughter?
e. She will have three sons?
2. Paul historically can make a free throw 72% of the time, what is the probability
a. He will make 8 out of his next 10 baskets?
b. He will make 4 out of his next 12 baskets?
c. He will make 2 out of his next 3 baskets?
d. He will make 7 out of 7 baskets?
e. He will make at least one of his next 5 baskets?
3. Dorian says “Um” 22% of the time. What is the probability
a. He will say “Um” twice in the next 9 words?
b. He will say “Um” four times in the next 20?
c. He won’t say “Um” in 30 words?
d. He will say “Um” at least twice in the next 15 words?
e. He will say “Um” 4 out of the next 5 words?
4. Jenna has a strange allergy. She’s allergic to 45% of all dogs. What is the probability
a. She will be allergic to 3 out of the next 8 dogs she encounters?
b. She will be allergic to 1 or 2 out of the next 12 dogs she encounters?
c. She will be allergic to 10 of the next 20 dogs she encounters?
d. She will be allergic to at least 1 of the next 5 dogs she encounters?
e. She will be allergic to all of the next 6 dogs she encounters?
263
Cogswell – PCM

Section
V.B.5 Mathematical
Expectation
In games of chance, there are often rewards for accomplish specific feats. The Mathematical
Expectation is how much on average each player is expected to win or lose. The games we might
play at carnivals are constructed in such a way that the probability of winning is extremely low enough
that the mathematical expectation for each person might be to lose. This way, even if some people
win, the game is expected to earn a profit if enough people play.
Example 1
Billie wants to play a game where he pays $2 to throw three balls at 9 fluffy-haired clown faces. If he
knocks down 1 face, he wins a 25-cent yo-yo, two faces will get him a $1.50 stuffed animal filled with
Styrofoam, and three faces will get him a $5 giant stuffed animal. However, the clown faces only take
up 25% of the area, the other 75% is made up of the deceptive fluffy hair. He must hit the face to
knock it down. Should he play the game?
Firstly, let’s look at the prizes. If he’s paying $2 to play, the 25 cent yo-yo is actually a loss of $1.75,
the $1.50 stuffed animal is a loss of 50 cents, and the $5 stuffed animal is only a net gain of $3. Don’t
forget if he misses all three throws, he loses his $2. Add on to that the fact that he only has a 25%
chance of knocking a face over, and we start to get a sneaky suspicion that this is a bad idea. We do
have to prove this mathematically, however.
situation is a binomial expansion where A = head knocked down and ~A = head not knocked down.
We need to calculate the four distinct possibilities:
P(A = 0) = 3 0 (. 25) (. 75) = .422
P(A = 1) = 3 1 (. 25) (. 75) = .422
P(A = 2) = 3 2 (. 25) (. 75) = .141
P(A = 3) = 3 3 (. 25) (. 75) = .016
Remember, if we add these together we should get 1. And if we look at the probabilities, only 1 or 2
people in 100 players will earn the big prize. To calculate the mathematical expectation of the game,
we multiply each probability by its financial outcome and add them together.
. 422(−2) + .422(−1.75) + .141(−.5) + .016(3) = −$1.61.
On average a person would expect to lose $1.61 playing this game. If 100 people play, even with the
1 or 2 that win the grand prize, the booth operator will expect to make $161. Pretty crooked, huh?
264
Cogswell – PCM

V.B.5 Independent Practice
Answer the following questions about the following carnival games.
1. DOORDINARY: You are presented with 7 doors. Behind 4 of them is a red square, behind 2
of them is a yellow square, and behind 1 of them is a green square. The doors randomize on
each go. For $2 a player can open a single door. If they find the green square, they get a $10
stuffed animal. If they find the yellow square, they get a $0.50 pen. And if they find the red
square, they get nothing.
a. P(Green) =
b. P(Yellow) =
c. P(Red) =
d. How much should a person expect to win or lose for each round?
e. How much should be won or lost total when 100 people play?
2. HOOP IT UP: For $3 you get to shoot 3 free throws. If you make all 3, you get a $15
basketball. If you make 2 out of the 3, you get a $4 mini basketball. If you make 1 out of the 3,
you get a $1 golf ball painted to look like a basketball. If you make no baskets, you win
nothing. The hoop, however, is bent, so there is only a 23% you can actually make the basket.
a. P(0) =
b. P(1) =
c. P(2) =
d. P(3) =
e. How much should a person expect to win or lose for each round?
f. How much should be won or lost total when 100 people play?
3. BEANBAG BULLSEYE: For $1, you can fire a bean bag out of a cannon at a giant target. If
you miss the target, you are allowed to throw again until you hit the target. The middle circle
has a radius of 1 foot, the middle ring has an outer radius of 3 feet, and the outer-most ring has
an outer radius of 6 feet. If the bean bag hits the bullseye, you will win a large $5 bean bag
animal. If the bean bag hits the middle ring, you will win a $1.50 bean bag animal. If you hit
the outer-most ring, you will win a $0.50 plastic baggie filled with beans.
a. P(Bullseye) =
b. P(Middle) =
c. P(Outer) =
d. How much should a person expect to win or lose for each round?
e. How much should be won or lost total when 100 people play?
265
Cogswell – PCM

266
Cogswell – PCM