April 3, 2014

Holt Algebra 1
9-3 Graphing Quadratic FunctionsApril 3, 2014
Today:
 2 STAR Tests Completed
 Warm-Up
 Getting to Know the Quadratic Function:
(how the a, b, & c values change the parabola)
 Class Work

Holt Algebra 1
9-3 Graphing Quadratic Functions
x = 0
x = 1
(0, 2)
1. y = 4x2 – 7
2. y = x2 – 3x + 1
Find the axis of symmetry.
3. y = –2x2 + 4x + 3
(2, -12)5. y = x2 + 4x + 5 6. y = -2x2 + 2x – 8
Find the vertex and state whether
the graph opens up or down.
Warm-Up

Holt Algebra 1
For a quadratic function in the form y = ax2 + bx + c,
when x = 0, y = c. The y-intercept of a quadratic function is c
Finding the Y intercept
Find the vertex and the y-intercept
1. y = x2 – 2 y = x2 – 4x + 4 y = -2x2 – 6x - 3

Holt Algebra 1
Effects of the a, b, & c values
With your graph paper, graph the function: y = x2
This is called the parent
function. All other
quadratic functions are
simply transformations of
the parent.
For the parent function f(x) = x2:
• The axis of symmetry is x =
0, or the y-axis.
• The vertex is (0, 0)
• The function has only one
zero, 0.

Holt Algebra 1
9-3 Graphing Quadratic FunctionsEffects of the a, b, & c values

Holt Algebra 1
The value of a in a quadratic function determines not
only the direction a parabola opens, but also the width
of the parabola.

Holt Algebra 1
Example 1A: Comparing Widths of Parabolas
Order the functions from narrowest graph to widest.
f(x) = 3x2, g(x) = 0.5x2, h(x) = 1.5x2
f(x) = 3x2
h(x) = 1.5x2
g(x) = 0.5x2
The function with the
narrowest graph has
the greatest |a|.

Holt Algebra 1

Holt Algebra 1
The value of c in a quadratic function determines not only
the value of the y-intercept but also a vertical translation of
the graph of f(x) = ax2 up or down the y-axis.

Holt Algebra 1
Comparing Graphs of Quadratic Functions
Compare the graph of the function with the graph of f(x) = x2
opens downward and the graph of
f(x) = x2 opens upward.
• The graph of
is wider than the graph of
• The graph of
f(x) = x2.

Holt Algebra 1
9-3 Graphing Quadratic FunctionsCompare the graph of each the graph of f(x) = x2.
g(x) = –x2 – 4
• The graph of g(x) = –x2 – 4
opens downward and the graph
of f(x) = x2 opens upward.
The vertex of g(x) = –x2 – 4
f(x) = x2 is (0, 0).
is translated 4 units down to (0, –3).
• The vertex of
• The axis of symmetry is the same.

Holt Algebra 1
Example 1: Graphing a Quadratic Function
Graph y = 3x2 – 6x + 1.
Step 1 Find the axis of symmetry.
= 1
The axis of symmetry is x = 1.
Simplify.
Use x = . Substitute 3
for a and –6 for b.
Step 2 Find the vertex.
y = 3x2 – 6x + 1
= 3(1)2 – 6(1) + 1
= 3 – 6 + 1
= –2
The vertex is (1, –2).
The x-coordinate of the vertex
is 1. Substitute 1 for x.
Simplify.
The y-coordinate is –2.

Holt Algebra 1
Example 1 Continued
Step 3 Find the y-intercept.
y = 3x2 – 6x + 1
y = 3x2 – 6x + 1
The y-intercept is 1; the graph passes through (0, 1).
Identify c.

Holt Algebra 1
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 = 1, choose x-
values less than 1.
Let x = –1.
y = 3(–1)2 – 6(–1) + 1
= 3 + 6 + 1
= 10
Let x = –2.
y = 3(–2)2 – 6(–2) + 1
= 12 + 12 + 1
= 25
Substitute
x-coordinates.
Simplify.
Two other points are (–1, 10) and (–2, 25).
Example 1 Continued

Holt Algebra 1
Graph y = 3x2 – 6x + 1.
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.
Example 1 Continued
x = 1(–2, 25)
(–1, 10)
(0, 1)
(1, –2)
x = 1
(–1, 10)
(0, 1)
(1, –2)
(–2, 25)

Holt Algebra 1
Because a parabola is symmetrical, each point is
the same number of units away from the axis of
symmetry as its reflected point.
Helpful Hint

Holt Algebra 1
Example 2
Graph the quadratic function.
y = 2x2 + 6x + 2
Simplify.
for a and 6 for b.
The axis of symmetry is x .

Holt Algebra 1
y = 2x2 + 6x + 2
Simplify.
Example 2 Continued
= 4 – 9 + 2
= –2
The x-coordinate of the vertex is
. Substitute for x.
The y-coordinate is .
The vertex is .

Holt Algebra 1
y = 2x2 + 6x + 2
y = 2x2 + 6x + 2
Identify c.
Example 2 Continued

Holt Algebra 1
the y-intercept.
Let x = –1
y = 2(–1)2 + 6(–1) + 1
= 2 – 6 + 2
= –2
Let x = 1
y = 2(1)2 + 6(1) + 2
= 2 + 6 + 2
= 10
Substitute
x-coordinates.
Simplify.
Two other points are (–1, –2) and (1, 10).
Example 2 Continued
Since the axis of symmetry is x = –1 , choose x
values greater than –1 .

Holt Algebra 1
two other points.
across the axis of
symmetry. Connect the
points with a smooth curve.
y = 2x2 + 6x + 2
Example 2 Continued
(–1, –2)
(1, 10)
(–1, –2)
(1, 10)

Holt Algebra 1
Class Work:
Graphing Quadratic Functions

Holt Algebra 1
Example 3
Graph the quadratic function.
y + 6x = x2 + 9
Simplify.
for a and –6 for b.
= 3
y = x2 – 6x + 9 Rewrite in standard form.

Holt Algebra 1
Simplify.
Example 3 Continued
= 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

Holt Algebra 1
y = x2 – 6x + 9
y = x2 – 6x + 9
Identify c.
Example 3 Continued

Holt Algebra 1
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).
Example 3 Continued

Holt Algebra 1
two other points.
across the axis of symmetry.
Connect the points with a
smooth curve.
y = x2 – 6x + 9
Example 3 Continued
x = 3
(3, 0)
(0, 9)
(2, 1)
(1, 4)
(0, 9)
(1, 4)
(2, 1)
x = 3
(3, 0)

Holt Algebra 1
Example 2: Application
The height in feet of a basketball that is
thrown can be modeled by f(x) = –16x2
+ 32x, where x is the time in seconds
after it is thrown. Find the basketball’s
maximum height and the time it takes
the basketball to reach this height. Then
find how long the basketball is in the
air.

Holt Algebra 1
Example 2 Continued
1 Understand the Problem
The answer includes three parts: the
maximum height, the time to reach the
maximum height, and 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:

Holt Algebra 1
2 Make a Plan
Find the vertex of the graph because the
maximum height of the basketball and the
time it takes to reach it are the coordinates of
the vertex. The basketball will hit the ground
when its height is 0, so find the zeros of the
function. You can do this by graphing.
Example 2 Continued

Holt Algebra 1
Solve3
Use x = . Substitute
–16 for a and 32 for b.
Simplify.
Example 2 Continued

Holt Algebra 1
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.
Example 2 Continued

Holt Algebra 1
Identify c.f(x) = –16x2 + 32x + 0
The y-intercept is 0; the graph passes
through (0, 0).
Example 2 Continued

Holt Algebra 1
Step 4 Graph the axis of symmetry, the
vertex, and the point containing the y-intercept.
Then reflect the point across the axis of
symmetry. Connect the points with a smooth
curve.
(0, 0)
(1, 16)
(2, 0)
Example 2 Continued

Holt Algebra 1
The vertex is (1, 16). So at 1 second, the
basketball has reached its maximum height of 16
feet. The graph shows the zeros 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.
Example 2 Continued
(0, 0)
(1, 16)
(2, 0)

Holt Algebra 1
Look Back4
Check by substitution (1, 16) and (2, 0)
into the function.
16 = 16
0 = 0
Example 2 Continued
16 = –16(1)2 + 32(1)
?
16 = –16 + 32
?
0 = –16(2)2 + 32(0)
?
0 = –64 + 64
?

Holt Algebra 1
The vertex is the highest or lowest point on a
parabola. Therefore, in the example, it gives the
maximum height of the basketball.
Remember!

Holt Algebra 1
Check It Out! Example 2
As Molly dives into her pool, her height
in feet above the water can be modeled
by the function f(x) = –16x2 +
24x, where x is the time in seconds after
she begins diving. Find the maximum
height of her dive and the time it takes
Molly to reach this height. Then find how
long it takes her to reach the pool.

Holt Algebra 1
1 Understand the Problem
The answer includes three parts: the
maximum height, the time to reach the
maximum height, and the time to reach the
pool.
Check It Out! Example 2 Continued
List the important information:
• The function f(x) = –16x2 + 24x models
the height of the dive after x seconds.

Holt Algebra 1
2 Make a Plan
Find the vertex of the graph because the
maximum height of the dive and the time it
takes to reach it are the coordinates of the
vertex. The diver will hit the water when its
height is 0, so find the zeros of the function.
You can do this by graphing.

Holt Algebra 1
Solve3
Use x = . Substitute
–16 for a and 24 for b.
Simplify.
The axis of symmetry is x = 0.75.

Holt Algebra 1
f(x) = –16x2 + 24x
= –16(0.75)2 + 24(0.75)
= –16(0.5625) + 18
= –9 + 18
= 9
The vertex is (0.75, 9).
Simplify.
The y-coordinate is 9.
The x-coordinate of
the vertex is 0.75.
Substitute 0.75 for x.

Holt Algebra 1
Identify c.f(x) = –16x2 + 24x + 0
The y-intercept is 0; the graph passes
through (0, 0).

Holt Algebra 1
Step 4 Find another point on the same side
of the axis of symmetry as the point
containing the y-intercept.
Since the axis of symmetry is x =
0.75, choose an x-value that is less than
0.75.
Let x = 0.5
f(x) = –16(0.5)2 + 24(0.5)
= –4 + 12
= 8
Another point is (0.5, 8).
Substitute 0.5 for x.
Simplify.

Holt Algebra 1
Step 5 Graph the axis of symmetry, the
vertex, the point containing the y-
intercept, and the other point. Then reflect the
points across the axis of symmetry. Connect
the points with a smooth curve.
(1.5, 0)
(0.75, 9)
(0, 0)
(0.5, 8) (1, 8)

Holt Algebra 1
The vertex is (0.75, 9). So at 0.75
seconds, Molly's dive has reached its maximum
height of 9 feet. The graph shows the zeros of
the function are 0 and 1.5. At 0 seconds the
dive has not begun, and at 1.5 seconds she
reaches the pool. Molly reaches the pool in 1.5
seconds.
(1.5, 0)
(0.75, 9)
(0, 0)
(0.5, 8) (1, 8)

Holt Algebra 1
Look Back4
Check by substitution (0.75, 9) and (1.5, 0)
into the function.
9 = 9 
0 = 0 
9 = –16(0.75)2 + 24(0.75)
?
9 = –9 + 18
?
0 = –16(1.5)2 + 24(1.5)
?
0 = –36 + 36
?

Holt Algebra 1
Lesson Quiz
1. Graph y = –2x2 – 8x + 4.
2. The height in feet of a
fireworks shell can be modeled
by h(t) = –16t2 + 224t, where
t is the time in seconds after it
is fired. Find the maximum
height of the shell, the time it
takes to reach its maximum
height, and length of time the
shell is in the air.
784 ft; 7 s; 14 s

Holt Algebra 1

April 3, 2014

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April 3, 2014