1. Today:
Review all Quadratics for Test Friday
New Solving & Graphing Method: Square Roots
Complete Class Work from Friday

2. 3rd Quarter Final Exam:
The exam is on Friday; be here.
10 Questions on:
A. Quadratics
B. Factoring ax2
+ bx + c polynomials
10 Questions on:
C. Exponents
D. Adding, Subtracting, Multiplying & Dividing
Monomials and Polynomials
Notebooks submitted Tues – Fri. (1st – 4th)

3. Warm-Up/Review:
1. y = 5x2
– 1; Graph the general shape and location from
the information given in the function. Width, shape,
vertex, y-intercept, etc. Even the missing ‘b’ term provides
information.

4. Warm-Up/Review:
Find the axis of symmetry for the graph y = 3x2
+ 6x + 4
a. y = -1 b. x = 1 c. x = -1 d. y = 1
A graph of a quadratic function has x intercepts of (3,0)
and (-7,0). Which of the following could match the graph?
a. x2
+ 4x - 21 = 0 b. x2
- 10x - 21 = 0
c. x2
+ 10x + 21 = 0 d. x2
- 4x + 21 = 0

5. Warm-Up/Review:

6. Warm-Up/Review:

7. Warm-Up/Review:(3)

8. Warm-Up/Review:
Why or why not?
Why or why not?

9. Warm-Up/Review:

11. Holt Algebra 1
9-3 Graphing Quadratic Functions
Write down the steps, in order, that you would follow to
solve the equation
y + 6x = x2 + 9
Step 2:
The axis of symmetry is x = 3.
= 3
y = x2 – 6x + 9Rewrite in standard form.Step 1
Find the axis of symmetry.

12. Holt Algebra 1
9-3 Graphing Quadratic Functions
Step 3:
= 9 – 18 + 9
= 0
The vertex is (3, 0).
The y-coordinate is 0. .
y = x2 – 6x + 9
y = 32 – 6(3) + 9
Find the vertex.

13. Holt Algebra 1
9-3 Graphing Quadratic Functions
Step 4:
y = x2 – 6x + 9
y = x2 – 6x + 9
The y-intercept is 9; the graph passes through (0, 9).
Identify c.
Find the y-intercept.

14. Holt Algebra 1
9-3 Graphing Quadratic Functions
Step 5:
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
Two other points are (2, 1) and (1, 4).
Find two more points on the same side of the axis of
symmetry as the point containing the y- intercept.

15. Holt Algebra 1
9-3 Graphing Quadratic Functions
Step 6:
Reflect the points across the axis of symmetry. Connect
the points with a smooth curve.
y = x2 – 6x + 9
x = 3
(3, 0)
(0, 9)
(2, 1)
(1, 4)
(0, 9)
(1, 4)
(2, 1)
x = 3
(3, 0)
Graph the axis of symmetry, the vertex, the point containing
the y-intercept, and two other points.
Step 7:

16. Holt Algebra 1
9-3 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.
*The next question will give you honest feedback about your
recent focus, attention, and work habits. At this point, we should
all be able to answer it. Write, do not say the answer, please.
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.

17. Holt Algebra 1
9-3 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)

18. Holt Algebra 1
9-3 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......

19. Holt Algebra 1
9-3 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.

20. Holt Algebra 1
9-3 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.

21. Holt Algebra 1
9-3 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).

22. Holt Algebra 1
9-3 Graphing Quadratic Functions
Step 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)

23. Holt Algebra 1
9-3 Graphing Quadratic Functions
The 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.

25. Holt Algebra 1
9-3 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!
Friday’s Class Work...Plus the following

26. Holt Algebra 1
9-3 Graphing Quadratic Functions

27. Holt Algebra 1
9-3 Graphing Quadratic Functions