10-7 Using Factors to Sketch y = x 2 + bx + c notes

Mrs. Aitken’s Integrated 1 Math
Unit 10 Quadratic Equations as Models
10-7 Using Factors to Sketch y = x 2 + bx + c
Warm-up
Find a pair of numbers with each sum and product.
1. a sum of 5 and a product of 6 2 and 3
2. a sum of -2 and a product of -35 -7 and 5
3. a sum of -8 and a product of 16 -4 and -4
4. (x – 3) (x – 7) x 2 – 10x + 21
5. Solve 8x + 4 = 2 x = -1/4
6. x 2 = 169 ±13
7. x 2 = 48 ±4√3
Today we will:
1. Factor trinomials and sketch the graph of a parabola.
Following class period we will:
1. Solve problems about situations modeled by parabolas. Last section in the
book!
10-7 Using Factors to Sketch y = x 2 + bx + c
Remember - expanded form of a trinomial is x 2 + bx + c.
There is a relationship that exists between b and c that is used to write the
trinomial in factored form.
x 2 + bx + c = (x + ? ) (x + ? )
These two integers, ?, must have a
product of c and a sum of b.
Example 1:
Factor x 2 - x – 6
Solution
c = -6
b = -1
Find a pair of integers whose product is -6.
10-7 Using Factors to Sketch y = x 2 + bx +c

Mrs. Aitken’s Integrated 1 Math
Unit 10 Quadratic Equations as Models
You can pick the order to guess and check…check using FOIL.
Factors of -6 Plug these in and use guess and check to figure out your answer.
Use FOIL to check your answer
-2 • 3 (x - 2) (x + 3) x 2 + 3x - 2x - 6 = x 2 + x - 6 nope
-3 • 2 (x - 3) (x + 2) x 2 + 2x - 3x - 6 = x 2 - x - 6 YES
-1 • 6 (x - 1) (x + 6) x 2 + 6x - x - 6 = x 2 + 5x - 6 nope
-6 • 1 (x - 6) (x + 1) x 2 + x - 6x - 6 = x 2 - 5x - 6 nope
(x – 3)(x + 2)
Example 2:
Factor x 2 + 7x – 18
Solution
c = -18
b = 7
You can pick the order to guess and check.
Factors of -18 Plug these in and use guess and check to figure out your answer.
Use FOIL to check your answer
-1 • 18 (x - 1) (x + 18) x 2 + 18x - x - 18 = x 2 + 17x - 18 nope
-18 • 1 (x - 18) (x + 1) x 2 + x - 18x - 18 = x 2 - 17x - 18 nope
-2 • 9 (x - 2) (x + 9) x 2 + 9x - 2x - 18 = x 2 +7x - 18 YES
-9 • 2 (x - 9) (x + 2) x 2 + 2x - 9x - 18 = x 2 - 7x - 18 nope
-3 • 6 (x - 3) (x + 6) x 2 + 6x - 3x - 18 = x 2 + 3x - 18 nope
-6 • 3 (x - 6) (x + 3) x 2 + 3x - 6x - 18 = x 2 - 3x - 18 nope
(x – 2)(x + 9)
Side-note: Notice -2(9) = -18

Mrs. Aitken’s Integrated 1 Math
Unit 10 Quadratic Equations as Models
Example 3:
Factor x 2 + 10x + 24
Solution
c = 24
b = 10
Factors of 24 Plug these in and use guess and check to figure out your answer.
Use FOIL to check your answer
6 • 4 (x + 6) (x + 4) x 2 + 6x +4x + 24 = x 2 + 10x + 24 YES
2 • 12 (x + 2) (x + 12) x 2 + 12x + 2x + 24 = x 2 + 14x + 24 nope
3 • 8 (x + 3) (x + 8) x 2 + 8x + 3x + 24 = x 2 +11x + 24 nope
1 • 24 (x + 1) (x + 24) x 2 + 24x + x + 24 = x 2 + 25x + 24 nope
-6 • -4 (x - 6) (x - 4) x 2 - 4x - 6x + 24 = x 2 - 10x + 24 nope
-2 • -12 (x - 2) (x - 12) x 2 - 12x - 2x + 24 = x 2 - 14x + 24 nope
-3 • -8 (x - 3) (x - 8) x 2 - 3x - 8x + 24 = x 2 - 11x + 24 nope
-1 • -24 (x - 1) (x - 24) x 2 - 24x - x + 24 = x 2 - 25x + 24 nope
(x + 6)(x + 4)
Example 4:
Factor x 2 - 5x + 10
Solution 4
Factors of 10
Plug these in and use guess and check to figure out your answer.
Use FOIL to check your answer
2 • 5 (x + 2) (x + 5) x 2 + 5x +2x + 10 = x 2 + 7x + 10 nope
1 • 10 (x + 1) (x + 10) x 2 + 10x + x + 10 = x 2 + 11x + 10 nope
-2 • -5 (x - 2) (x - 5) x 2 - 5x + 2x + 10 = x 2 - 3x + 10 nope
-1 • -10 (x - 1) (x - 10) x 2 - 10x - x + 10 = x 2 - 11x + 10 nope
Unfactorable
* Sometimes a trinomial cannot be factored.
10-7 Using Factors to Sketch y = x 2 + bx +c

Mrs. Aitken’s Integrated 1 Math
Unit 10 Quadratic Equations as Models
Sketching the graph of y = x 2 + bx + c
You can make a reasonably accurate sketch of a parabola by determining the
line of symmetry, coordinated of the vertex, and the x- and y-intercepts of the
graph from the equation of the parabola.
Example 5: - Use the vertex, the intercepts, and symmetry to sketch the graph
of y = x 2 - 2x – 3.
Solution
Step 1 – find the equation of the line of symmetry.
x = -b/2a (Formula for the line of symmetry)
b = -2
a = 1
x = -(-2)/2(1) = 1
The equation of the line of symmetry is x = 1.
Step 2 - Find the coordinates of the vertex.
From the line of symmetry found in step 1, the x-coordinate of the vertex is 1.
Substitute 1 for x into the equation.
y = x 2 - 2x – 3
y = 1 2 – 2(1) – 3
y = 1 – 2 – 3
y = -4
The vertex is (1, -4).
Step 3 – Find the intercepts
a. Find the y-intercept. Set x = 0 and solve for y.
y = 0 2 – 2(0) – 3
y = -3
(0, -3)
b. Find the x-intercepts
Factor the right side of the equation.
y = x 2 - 2x – 3
y = (x – 3)(x + 1)
x – 3 = 0 x + 1 = 0
x = 3 x = -1
(3, 0) (-1, 0)
Use the zero product property to set
each factor to 0 and solve.
10-7 Using Factors to Sketch y = x 2 + bx +c

Mrs. Aitken’s Integrated 1 Math
Unit 10 Quadratic Equations as Models
Step 4 – Use the vertex, intercepts, and symmetry to sketch the parabola.
By symmetry, the point (2, -3)
is on the parabola.
10-7 Using Factors to Sketch y = x 2 + bx +c