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

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