This document provides instructions for graphing quadratic functions and examples worked through step-by-step. It begins with the general steps: 1) identify coefficients, 2) find the vertex, 3) draw the axis of symmetry, 4) find the y-intercept, 5) find roots, 6) reflect points over the axis, and 7) graph the parabola. An example graphs the function y = 3x^2 - 6x + 1. It then works through graphing the path of a basketball using the function f(x) = -16x^2 + 32x, finding that the maximum height is 16 feet reached at 1 second, and the basketball is in the air for 2 seconds. The document

1. Today:
 Review Khan Academy Topics
 Graphing Quadratic Functions:
 Using Square Roots to Solve Equations
 Class Work 4.2

5. STEP 1 Identify the coefficients of the function.
STEP 2
Find the vertex. Calculate the x - coordinate.STEP 3
Draw the axis of symmetry
STEP 4 Identify the y - intercept c,
STEP 5 Find the roots by using one of the solution methods,
We are unable to find the roots with our knowledge for now, so
we'll select another value of x and solve for y. The AOS is 1, so let's
choose x = -1. Find the y coordinate.
The two other points are (–1, 10) and (–2, 25)
STEP 6 Reflect this point over the AOS to plot another point.
STEP 7 Graph the parabola
Graph a function of the form y = ax2 + bx + c
y = 3x2 – 6x + 1, Plot 5 points and draw the curveGraph

6. x = 1
(–1, 10)
(0, 1)
(1, –2)
(–2,
25)

7. Graph a function of the form y = ax2 + bx + c
Step 1: Find the axis of symmetry.
Use x = . Substitute 1 for a
and –6 for b.
The axis of symmetry is x = 3.
= 3
y = x 2 – 6x + 9 Rewrite in standard form.
y + 6x = x2 + 9Graph the quadratic function

8. Step 2: Find the vertex.
= 9 – 18 + 9 = 0
The vertex is (3, 0).
The x-coordinate of the vertex is 3.
Substitute 3 for x.
The y-coordinate is 0.
y = x2 – 6x + 9
y = 32 – 6(3) + 9
Graph a function of the form y = ax2 + bx + c
y = x 2 – 6x + 9

9. Step 3: Find the y-intercept.
y = x2 – 6x + 9
y = x2 – 6x + 9
The y-intercept is 9; the graph passes through (0, 9).
Identify c.
Graph a function of the form y = ax2 + bx + c
y = x 2 – 6x + 9

10. Step 4 Find two more points on the same side of the axis of
symmetry as the point containing the y- intercept.
Since the axis of symmetry is x = 3, choose x-values less
than 3.
Let x = 2
y = 1(2)2 – 6(2) + 9
= 4 – 12 + 9
= 1
Let x = 1
y = 1(1)2 – 6(1) + 9
= 1 – 6 + 9
= 4
Substitute
x-coordinates.
Simplify.
Two other points are (2, 1) and (1, 4).
Graph a function of the form y = ax2 + bx + c
y = x 2 – 6x + 9

11. Step 5 Graph the axis of symmetry,
the vertex, the point containing the
y-intercept, and two other points.
Step 6 Reflect the points across
the axis of symmetry. Connect
the points with a smooth curve.
y = x 2 – 6x + 9
x = 3
(3, 0)
(0, 9)
(2, 1)
(1, 4)
(6, 9)
(5, 4)
(4, 1)
x = 3
(3, 0)

12. Graphing Quadratic Functions
After a player takes a shot, the height in feet of a basketball can be
modeled by f(x) = –16x2 + 32x, where 'x ' is the time in seconds after
it is shot. Find
1. The basketball’s maximum height
2. The time it takes the basketball to reach this height.
3. How long the basketball is in the air.
There is no c term in this equation. What does
that tell us about our graph??
The graph is not shifted up or down the y axis, therefore
the y-intercept is at the origin, which also means one of
the solutions must be zero.
Quadratic Applications:
Write, do not say the answer, please.

13. Graphing Quadratic Functions
1 Understand the Problem
Our answer includes three parts:
1. The maximum height of the ball,
2. The time to reach the maximum height, and
3. The time to reach the ground.
• The function f(x) = –16x2 + 32x models the height of
the basketball after x seconds.
List the important information:
What are the two variables for our x and y axes.
(Plural of axis, pronounced ax-eez)

14. Graphing Quadratic Functions
2 Make a Plan
The basketball will hit the ground when its height is 0.
Round to the nearest whole number if necessary.
What parts of the graph are important in solving our problem?
A. The vertex. Why?
A. Because the maximum height of the basketball and the
time it takes to reach it are the coordinates of the vertex.
B. The zero's of the function because......

15. Graphing Quadratic Functions
Solve3
Step 1 Find the axis of symmetry.
Use x = . Substitute
–16 for a and 32 for b.
Simplify.
The axis of symmetry is x = 1.

16. Graphing Quadratic Functions
Step 2 Find the vertex.
f(x) = –16x2 + 32x
= –16(1)2 + 32(1)
= –16(1) + 32
= –16 + 32 = 16
The vertex is (1, 16).
The x-coordinate of the
vertex is 1. Substitute
1 for x.
Simplify.
The y-coordinate is 16.

17. Graphing Quadratic Functions
Step 3 Find the y-intercept.
Identify c.f(x) = –16x2 + 32x + 0
The y-intercept is 0; the graph passes through (0, 0).

18. Graphing Quadratic FunctionsStep 4: Graph the axis of symmetry, the vertex, and the point
containing the y-intercept. Then use symmetry to reflect the point
across the axis of symmetry. Connect the points with a smooth
curve.
(0, 0)
(1, 16)
(2, 0)

19. Graphing Quadratic FunctionsThe vertex is (1, 16). So at 1 second, the basketball has reached its
maximum height of 16 feet.
(0, 0)
(1, 16)
(2, 0)
The graph shows the zero’s of the function are 0 and 2. At 0 seconds
the basketball has not yet been thrown, and at 2 seconds it reaches the
ground. The basketball is in the air for 2 seconds.

21. Graphing Quadratic Functions
The vertex is the highest or lowest point on a parabola.
Therefore, it always represents the maximum height of an
object following a parabolic path.
Remember!
Tuesday's Class Work...Plus the following

22. Graphing Quadratic Functions
Class Work 4.2
~Find the important points on the graph. Label them, and draw the
curve.
Show your work.

23. Graphing Quadratic Functions

24. Graphing Quadratic Functions