3.1
Name
Quadratic Functions in
Vertex Form
Class
3.1
Essential Question: What does the vertex form of a quadratic function reveal
about the function?
Texas Math Standards
A2.4.D: Transform a quadratic function ƒ(x) = ax 2 + bx + c to the form ƒ(x) = a(x - h) + k
to identify the different attributes of ƒ(x).
Transform a quadratic function f(x) = ax 2 + bx + c to the
form f(x) = a(x – h) 2 + k to identify the different attributes of f(x).
(
A2.1.F
Remember that when a > 0, the graph of ƒ(x) = ax 2 opens upward, so the graph of g(x) = -ax 2,
which is the reflection of ƒ(x) in the x-axis, opens downward. In general, the graph of a quadratic
function in vertex form opens upward when a > 0 and downward when a < 0, as illustrated by
the graphs.
The student is expected to analyze mathematical relationships to connect
and communicate mathematical ideas.
Language Objective
f(x)
4
1.C.4, 1.E.1, 1.E.2, 2.C.2, 5.B.2
Work with a partner to translate the variables in a quadratic function into
words.
f(x) (h, k) = (2, 4)
4
2
2
x
-4
ENGAGE
-2
0
x
-4 -2 0
a = -2 -2
2
4
a = 0.5
-2
© Houghton Mifflin Harcourt Publishing Company
-4
(h, k) = (-1, -3)
ƒ(x) = 0.5(x-(-1)) - 3
2
For the function in vertex form ƒ(x) = 0.75(x + 3) - 2, h =
-3
2
f(x)= 0.75(x + 3) 2 - 2
x
f(x)= 0.75(x + 3) 2 - 2
x
-7
f(h - 4) =
10
h+1=
-2
f(h + 1) = -1.25
h-2=
-5
f(h - 2) =
1
h+2=
-1
f(h + 2) =
1
h-1=
-4
f(h - 1) = -1.25
h+4=
1
f(h + 4) =
10
h=
-3
Module 3
f(h) =
-2
ges
EDIT--Chan
DO NOT Key=TX-A
Correction
must be
Lesson 1
117
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Date
Class
Functions
QuadraticForm
in Vertex
Name
3.1
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PAGES 87–98
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ƒ(x) = 0.5

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2
0
-4 -2
a = -2 -2
x
4
2
a = 0.5
0
-2
-4
-4
-3)
(h, k) = (-1,
2
4
(x - 2) +
ƒ(x) = -2
-3
on in vertex
For the functi
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h-1=
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= 0.75(x
2
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A2_MTXESE353930_U2M03L1.indd 117
g Attribut
Investigatin
a
Form
h) + k is
__1
in Vertex
⋅ ƒ( b(x - ) b, h, and k
(x) = a
-7
-5
-4
-3
f(h - 4) =
f(h - 2) =
f(h - 1) =
f(h) =
10
-3
. Complete
h+2=
1
h+4=
-1.25
-2
the table.
2
2
(x + 3) -
f(x)= 0.75
x
h+1=
-2
-1
1
f(h + 1) =
-1.25
f(h + 2) =
f(h + 4) =
1
10
Lesson 1
117
L1.indd
0_U2M03
SE35393
A2_MTXE
Lesson 3.1
. Complete the table.
h-4=
Module 3
117
4
ƒ(x) = -2(x - 2) + 4
2

2
-4
n Mifflin
View the online Engage. Discuss the forces involved
in determining the height of a volleyball hit upwards,
what makes it go up, and what makes it come back
down. Introduce the concept of projectile motion and
how such motion can be described mathematically.
Then preview the Lesson Performance Task.
)
You have seen before that the graph of the function g(x) = a ⋅ ƒ __b1 (x - h) + k is a
transformation of the graph of a parent function ƒ(x), with the parameters a, b, h, and k
indicating the specific transformations. For the parent quadratic function ƒ(x) = x 2 and b = 1,
2
the equation of the transformation is ƒ(x) = a(x - h) + k. This is called the vertex form of a
quadratic function since the vertex of the graph is located at (h, k).
Mathematical Processes
PREVIEW: LESSON
PERFORMANCE TASK
Investigating Attributes of a Quadratic Function
in Vertex Form
Explore
A2.4.D
Possible answer: It shows whether the function has
a maximum or a minimum and its value, the x-value
where it occurs, range, and intervals where it is
increasing or decreasing. For the graph, it indicates
direction of opening, vertex, axis of symmetry, end
behavior, and whether it is a vertical stretch or
compression of the parent quadratic graph.
Resource
Locker
2
The student is expected to:
Essential Question: What does the
vertex form of a quadratic function
reveal about the function?
Date
Quadratic Functions
in Vertex Form
© Houghto
LESSON
117
2/22/14
4:52 AM
2/22/14 4:52 AM
What do you observe about the values of ƒ(x) for values of x to either side of x = h?
EXPLORE
The value of f(x) is the same for any two values of x that are the same distance to
either side of x = h.
Investigating Attributes of a
Quadratic Function in Vertex Form
What does your observation say about the axis of symmetry of the graph?
The graph is symmetric about the line x = h, the vertical line through the vertex.
So, the line x = h is the axis of symmetry of the graph.
B
QUESTIONING STRATEGIES
In Step A, you saw that for a quadratic function in vertex form, ƒ(h - 1) = ƒ(h + 1),
ƒ(h - 2) = ƒ(h + 2), and so on.
For ƒ(x) = 3(x + 2) - 1: a =
2
-2
,h=
3
,k=
What does the sign of a tell you about the
value of k? If a is a positive number, then k is
the minimum value of the function. If it is a negative
number, then k is the maximum value of the
function.
-1
Complete the table.
f(x) = 3(x + 2) 2 - 1
(
f( h ) = 3
) -1=
2
0
(
f(h - 2) = f(h + 2) = 3(±
f(h - 3) = f(h + 3) = 3(±
f(h - 4) = f(h + 4) = 3(±
f(h - 1) = f(h + 1) = 3 ±
1
2
3
4
-1
) -1=
) -1=
) -1=
) -1=
2
2
2
2
Why do you think the graph of
2
f(x) = a(x - h) + k is symmetric about the
line x = h? The value x – h, the distance of x from h,
is squared. There are two values of x that are the
same distance from h that result in the same value
2
of (x – h) and thus have the same function value.
2
11
26
47
© Houghton Mifflin Harcourt Publishing Company
Module 3
118
Lesson 1
PROFESSIONAL DEVELOPMENT
A2_MTXESE353930_U2M03L1.indd 118
Learning Progressions
2/20/14 12:13 AM
Students were exposed to the vertex form of a quadratic function,
2
f(x) = a(x - h) + k, in Module 1, where they learned to identify how the
parameters a, h, and k produced transformations of the graph of the parent
quadratic function f(x) = x 2. In this lesson, students learn how to convert
quadratic functions to vertex form, and how they can use the result to identify the
attributes of the function. Work with vertex form will continue in the following
lesson, where students will learn how to write quadratic functions in vertex form
given information about the graph of the function.
Quadratic Functions in Vertex Form 118
When x = h, (x + 2) =
2
0 , 3(x + 2) 2=
0 , and the value of ƒ(x) is
[greater than/less than/equal to] the parameter k.
When x ≠ h, (x + 2) is [positive/negative,] a(x + 2) = 3(x + 2) is [positive/negative,]
2
2
2
and the value of ƒ(x) is [greater than/less than/equal to] the parameter k.
So, a is [positive/negative,] and the function has a [maximum/minimum] value of k
when x = h.
For ƒ(x) = -2(x - 6) + 4: a = -2 , h =
2
6 ,k=
4
Complete the table.
f(x) = -2(x - 6) + 4
2
(
f(h) = -2
) +4=
2
0
(
f(h - 2) = f(h + 2) = -2(±
f(h - 3) = f(h + 3) = -2(±
f(h - 4) = f(h + 4) = -2(±
f(h - 1) = f(h + 1) = -2 ±
When x = h, (x - 6) =
2
1
2
3
4
4
) +4= 2
) + 4 = -4
) + 4 = -14
) + 4 = -28
2
2
2
2
0 , -2(x - 6) 2 =
0 , and the value of ƒ(x) is
© Houghton Mifflin Harcourt Publishing Company
[greater than/less than/equal to] the parameter k.
When x ≠ h, (x - 6) is [positive/negative], a(x - 6) = -2(x - 6) is [positive/negative],
and the value of ƒ(x) is [greater than/less than/equal to] the parameter k.
2
2
2
So, a is [positive/negative], and the function has a [maximum/minimum] value of k when x = h.
Reflect
1.
Find ƒ(h + d) and ƒ(h - d) for a general quadratic function in vertex form, ƒ(x) = a(x - h) + k. How
does this verify your observation
in Step A? Do the values of a and k affect this result?
2
f(h + d) = a (h + d) - h + k = ad 2 + k;
2
(
)
2
f(h - d) = a((h - d) - h) + k = a(-d) + k = ad 2 + k;
2
Because both f(h + d) and f(h - d) equal ad 2 + k for any value of d, the value of
the function is the same for any pair of domain values that are the same distance to
either side of x = h. The values of a and k do not affect this result.
Module 3
119
Lesson 1
COLLABORATIVE LEARNING
A2_MTXESE353930_U2M03L1.indd 119
Peer-to-Peer Activity
Have students work in pairs. Provide each pair of students with a quadratic
function written in standard form. Have one student provide verbal instructions
for converting the function to vertex form to the partner, while the partner
performs the steps. Once the function is written in vertex form, ask students to
identify the attributes of the function. Then have partners switch roles, repeating
the activity for a different function.
119
Lesson 3.1
2/20/14 12:13 AM
2.
What are the roles of the parameters a, h, and k in determining the maximum or minimum value of a
2
quadratic function ƒ(x) = a(x - h) + k?
The maximum or minimum value is k, and occurs when x = h. When a is positive,
EXPLAIN 1
the function has a minimum. When a is negative, the function has a maximum.
3.
Identifying the Attributes of a
Quadratic Function in Vertex Form
Discussion How can you find the range of ƒ(x) = a(x - h) + k without computing any function
values?
Since f(h) = k is a minimum when a > 0 and a maximum when a < 0, you can determine
2
the range from just a and k. When a > 0, the range of f(x) = a(x - h) + k is all values
2
AVOID COMMON ERRORS
greater than or equal to k. When a < 0, the range of f(x) = a(x - h) + k is all values less
2
Some students may make errors in identifying the
coordinates of the vertex of the parabola. Encourage
them to circle the value of h with the sign preceding
it, and to draw a square around the value of k and its
sign. Then tell them to think about the circle as a
letter o for opposite, and to use the opposite of that
value for the first coordinate in the vertex. The
second coordinate is the value of k.
than or equal to k.
4.
Discussion Describe the behavior of the graph of ƒ(x) = a(x - h) + k on either side of x = h when
a > 0 and when a < 0.
Since f(h) = k is a minimum when a > 0, the graph decreases toward k to the left of x = h
2
and increases away from k to the right. Similarly, since f(h) = k is a maximum when a < 0,
the graph increases toward k to the left of x = h and decreases away from k to the right.
Explain 1
Identifying the Attributes of a Quadratic Function
in Vertex Form
Attributes of a quadratic function ƒ(x) = a(x - h) + k
2
The graph has vertex ( h, k ) and axis of symmetry x = h.
When a > 0:
x= h
• ƒ(x) has a minimum value of k when x = h.
⎧
⎫
• The domain of ƒ(x) is ⎨x| x ∈ℜ⎬.
⎩
⎭
⎧
⎫
• The range of ƒ(x) is ⎨y| y ≥ k⎬.
⎩
⎭
• ƒ(x) is decreasing on (-∞, h) and increasing
on (h, + ∞).
When a < 0:
(h, k)
• ƒ(x) has a maximum value of k when x = h.
⎧
⎫
• The domain of ƒ(x) is ⎨x| x ∈ℜ⎬.
⎩
⎭
⎧
⎫
• The range of ƒ(x) is ⎨y| y ≤ k⎬.
⎩
⎭
• ƒ(x) is increasing on (-∞, h) and decreasing
on (h, + ∞).
Module 3
© Houghton Mifflin Harcourt Publishing Company
(h, k)
x= h
120
Lesson 1
DIFFERENTIATE INSTRUCTION
A2_MTXESE353930_U2M03L1.indd 120
Manipulatives
2/22/14 5:00 AM
Students can use algebra tiles to model x 2 + 6x + 4 , with 3 x-tiles to the right of
the x 2-tile and 3 x-tiles below the x 2-tile. Once the tiles are laid out, students should
move the 1-tiles, as a group, off to the side. Students then add enough positive 1-tiles
to the x 2-tile and 6 x-tiles to complete the square (9 tiles). They add an equal number
of negative 1-tiles to their separated group of 4 1-tiles so that the net amount added is
0. Combining the 1-tiles off to the side gives 5 negative 1-tiles, or -5. Students then
write the expression as a square of a binomial, (x + 3) 2, plus the number of 1-tiles off
to the side, (x + 3) 2 - 5, which is equal to the original trinomial.
Quadratic Functions in Vertex Form 120
Example 1
QUESTIONING STRATEGIES
Is it possible for a quadratic function to be
neither increasing nor decreasing over an
interval in its domain? Explain. No. The graph of a
quadratic function is a parabola, which is always
increasing on one side of its vertex and decreasing
on the other.

Identify the axis of symmetry of the graph of the function. Then give the
function’s maximum or minimum value, domain, range, and the intervals
where the function is increasing or decreasing.
2
ƒ(x) = 4(x - 3) + 5
2
Find a, h, and k in the vertex form ƒ(x) = a(x - h) + k.
ƒ(x) = 4(x - 3) + 5, so a = 4, h = 3, and k = 5.
2
Axis of symmetry: x = h, so x = 3
Since a > 0, ƒ(x) has a minimum value of k, or 5.
⎧
⎧
⎧
⎫
⎫
⎫
Domain: ⎨x| x ∈ℜ⎬; Range: ⎨y| y ≥ k⎬, so ⎨y| y ≥ 5⎬
⎩
⎩
⎩
⎭
⎭
⎭
Decreasing on (-∞, h), so decreasing on (-∞, 3); Increasing on (h, +∞), so increasing on (3, +∞)
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Math Connections

ƒ(x) = -11(x + 5) - 1.6
2
Find a, h, and k in the vertex form ƒ(x) = a(x - h) + k.
2
Discuss with students how the shape of the graph of a
quadratic function (and therefore its attributes) is
related to the fact that the rule for the function
contains x 2. Compare this to the graph of a linear
function, where there is no squared term.
( ( -5 )) + ( -1.6 ),
2
ƒ(x) = -11(x + 5) 2 - 1.6 = -11 x -
so a = -11 , h = -5 , and k = -1.6 .
Axis of symmetry: x = h, so x = -5
Since a < 0, ƒ(x) has a maximum value of k, or -1.6 .
⎧
⎧
⎫
Domain: ⎨x| x ∈ ℜ⎬; Range: ⎨y| y
⎩
⎩
⎭
≤
⎧
⎫
k⎬, so ⎨y| y
⎩
⎭
(
)
≤
⎫
⎭
-1.6⎬
Increasing on (-∞, h), so increasing on -∞, -5 ; Decreasing on (h, + ∞), so decreasing
© Houghton Mifflin Harcourt Publishing Company
on
( -5 , + )
∞
Your Turn
Identify the axis of symmetry. Then give the function’s maximum or minimum value,
domain, range, and the intervals where the function is increasing or decreasing.
2
1 (x + 1) 2 + 1
5. ƒ(x) = -(x - 0.8)
6. ƒ(x) = _
4
1
a = -1, h = 0.8, and k = 0
a = , h = -1, and k = 1
4
Axis of symmetry: x = -1
Axis of symmetry: x = 0.8
_
Maximum value of 0
⎧
⎧
⎫
⎫
D: ⎨x| x ∈ ℜ⎬; R: ⎨y| y ≤ 0⎬
⎩
⎩
⎭
⎭
Increasing on (-∞, 0.8]
Minimum value of 1
⎧
⎧
⎫
⎫
D: ⎨x| x ∈ ℜ⎬; R: ⎨y| y ≥ 1⎬
⎩
⎩
⎭
⎭
Decreasing on (-∞, -1]
Decreasing on [0.8, +∞)
Increasing on [-1, +∞)
Module 3
121
Lesson 1
LANGUAGE SUPPORT
A2_MTXESE353930_U2M03L1.indd 121
Communicate Math
Give each pair of students a note card with a quadratic function written in the
2
form f(x) = a(x - h) + k. Have students discuss what the variables a, h, and k
represent in the function, then create a “Quadratic Function in Vertex Form
Chart,” with written explanations for each variable. For example, a indicates a
reflection across the x-axis and/or a vertical stretch or compression; k represents a
vertical translation, and so on.
121
Lesson 3.1
2/22/14 5:07 AM
Explain 2
Converting from Standard Form to Vertex Form
EXPLAIN 2
Some attributes of a quadratic function are not easy to recognize from the standard form
ƒ(x) = ax 2 + bx + c. By completing the square, you can write an equivalent equation in vertex
form to make the attributes more apparent. Remember that a perfect square trinomial is the
square of a binomial.
Examples of perfect square trinomials:
x 2 + 12x + 36 = (x + 6)
(
9 = x-_
3
x 2 - 3x + _
4
2
2
)
2
Converting from Standard Form to
Vertex Form
x 2 + 2bx + b 2 = (x + b)
2
Note that the first and last terms of the trinomials are the squares of the first and last terms of the
binomials. The binomial’s last term is also half the coefficient of the trinomial’s middle term.
QUESTIONING STRATEGIES
Why does adding the square of half the
coefficient of x produce a perfect square
trinomial? Because perfect square trinomials are of
2
the form x 2 + 2kx + k 2, factorable as (x + k) .
Rewriting a Quadratic Function ƒ(x) = ax 2 + bx + c by Completing the Square
(
)
1.
If a ≠ 1, factor the first two terms as a x 2 + __ab x so that the coefficient of x 2 is 1.
2.
Set up to complete the square. The goal is to add a constant term inside the parentheses
to form a perfect-square trinomial. Because you are adding a quantity inside the
parentheses, you must subtract an equivalent quantity outside the parentheses to keep
the expression equivalent.
3.
b
b
inside the parentheses. Because ( __
is multiplied
Complete the square by adding ( __
2a )
2a )
2
( )
2
AVOID COMMON ERRORS
2
b
outside the parentheses.
by the coefficient a, subtract a __
2a
4.
Example 2

Write the quadratic function in vertex form by completing the square.
Then identify the maximum or minimum value of the function and the
value of x at which it occurs.
ƒ(x) = 2x 2 + 12x + 10
Factor so that the coefficient of x 2 is 1.
ƒ(x) = 2(x 2 + 6x) + 10
Set up to complete the square.
ƒ(x) = 2(x 2 + 6x +
(
( ) ) + 10 -2(_26 )
6
ƒ(x) = 2 x 2 + 6x + _
2
the square. Since this quantity is multiplied by 2,
subtract twice the quantity to keep the expression
equivalent.
2
© Houghton Mifflin Harcourt Publishing Company
b
_
a = 6; add the square of one half of 6 to complete
) + 10 -
2
ƒ(x) = 2(x 2 + 6x + 9) - 8
Simplify.
ƒ(x) = 2(x + 3) 2 - 8
Write the trinomial as the square of a binomial.
ƒ(x) = 2(x + 3) 2 - 8 = 2(x - (-3)) + (-8) , so a = 2, h = -3, and k = -8.
2
The function ƒ(x) has a minimum value of -8 when x = -3.
Module 3
A2_MTXESE353930_U2M03L1.indd 122
122
()
b is
Students may erroneously think that because __
2
always added to create the perfect square trinomial,
the same value (or a positive multiple of that value)
must be subtracted from c to maintain the equality.
Help them to see that this may not necessarily be the
case, pointing out that if a is a negative number, a
positive quantity is added to c.
Simplify, and write the trinomial as the square of a binomial.
Lesson 1
1/10/15 4:04 PM
Quadratic Functions in Vertex Form 122
B
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Reasoning
Discuss with students how the completing the square
process neither adds nor subtracts value to the
function rule. Students who have used the process in
the past to solve equations can compare that process
to the one used here, where quantities are added and
subtracted to the rule in order to maintain the
equality (as opposed to being added to both sides of
an equation).
ƒ(x) = -3x 2 + 24x - 43
Set up to complete the square.
ƒ(x) = -3 x 2 - 8x +
b
_
a = -8; add the square of one half of -8
-8
ƒ(x) = -3 x 2 - 8x + _
2
(
) - 43 -
( ( ))
2
-8
( ) (_
2 )
- 43 - -3
2
(
)
ƒ x = -3(x -4 ) + 5
ƒ(x) = -3 x 2 - 8x + 16 + 5
Simplify.
2
( )
Write the trinomial as the square of a binomial.
ƒ(x) = -3(x - 4) + 5, so a = -3 , h = 4 , and k = 5 .
2
The function ƒ(x) has a [maximum/minimum] value of 5 when x = 4 .
A graphing calculator can be used to check
that the function written in vertex form is
equivalent to the function written in standard form
by checking that the graphs of the two functions are
identical.
Reflect
7.
The introduction to this Explain began the process of completing the square for a general quadratic
function in standard form, ƒ(x) = ax 2 + bx + c. Complete this process:
ƒ(x) = ax 2 + bx + c
(
)
b
ƒ(x) = a x 2 + _
ax + c
(
© Houghton Mifflin Harcourt Publishing Company
2
(
(
)
b
_
2a
2
2
b
+c-a _
2a
( ) )+ c - _____
4a
b
b
_
ƒ(x) = a x 2 + _
a x + 2a
ƒ(x) = a x +
( )
( ))
b
b
_
ƒ(x) = a x + _
a x + 2a
b2
2
2
b
+c-_
4a
2
Explain how this result indicates how you can use the standard form to find the x-coordinate where the
function reaches its maximum or minimum value.
b
Because f(x) is now in vertex form, you can see that h = - ___
. So, the function reaches its
2a
b
maximum or minimum value (depending on the sign of a) when x = - ___
. You can read the
2a
values of a and b from the equation in standard form.
Module 3
A2_MTXESE353930_U2M03L1.indd 123
Lesson 3.1
)
ƒ(x) = -3 x 2 -8 x - 43
to complete the square. Since this quantity is
multiplied by -3, subtract -3 times the
quantity to keep the expression equivalent.
INTEGRATE TECHNOLOGY
123
(
Factor so that the coefficient of x 2 is 1.
123
Lesson 1
2/20/14 12:13 AM
Your Turn
EXPLAIN 3
Write the quadratic function in vertex form by completing the square. Then identify
the maximum or minimum value of the function and the value of x at which it occurs.
8.
Analyzing a Projectile Motion Model
ƒ(x) = -2x 2 - 8x + 3
f(x) = -2(x 2 + 4x)+ 3
4
f(x) = -2 x 2 + 4x +
2
(
(_) ) + 3 - (-2)(_42 )
2
2
QUESTIONING STRATEGIES
f(x) = -2(x + 2) + 11 = -2(x - (-2)) + 11,
2
2
Under what conditions is the time at which a
projectile reaches its maximum height exactly
one-half of the number of seconds the projectile is in
flight? This occurs when the initial height and the
final height of the projectile are the same.
a = -2, h = -2, and k = 11; value: 11 when x = -2
Explain 3
Analyzing a Projectile Motion Model
When you throw, hit, or launch an object into the air you can model its height under the influence of gravity as a
function of time. A model for projectile motion on Earth is h(t) = -16t 2 + v 0t + h 0, with height h in feet and time
t in seconds. In the model, “-16” is a constant that gives the downward acceleration from Earth’s gravity in feet per
second squared. The constant v 0 is the object’s initial vertical velocity in feet per second. The constant h 0 is the object’s
initial height (at time t = 0) in feet.
Example 3

Use the projectile motion model to answer the questions.
A golf ball is hit with an initial upward velocity of 64 feet per second off a
6-foot-high platform at a driving range.
a. How long does it take the ball to reach its maximum height? What is
that height?
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©cappi
thompson/Shutterstock
Use the model h(t) = -16t 2 + v 0 t + h 0 with v 0 = 64 and h 0 = 6:
h(t) = -16t 2 + 64t + 6
64 = 2
b = -_
Time at maximum height: t = -_
2a
2(-16)
Substitute t = 2 to find the maximum height:
h(2) = -16(2) + 64(2) + 6 = 70
2
The ball reaches its maximum height of 70 feet 2 seconds after it is hit.
Check your answer by graphing y = -16x 2 + 64x + 6 on a graphing
calculator and finding the coordinates of the maximum point.
Module 3
A2_MTXESE353930_U2M03L1.indd 124
124
Lesson 1
1/10/15 4:37 PM
Quadratic Functions in Vertex Form 124
b. What is the average rate of change in the ball’s height over the time interval from
when it is hit to when it reaches its maximum height? Over the time interval from
when it is at its maximum height to when it returns to the same height from which
it was hit?
AVOID COMMON ERRORS
Some students may think that the horizontal axis
of the graph of a function that models projectile
motion is used to indicate the horizontal distance
flown by the projectile, and thus the graph represents
its path. Help them to see that although the vertical
axis indicates the vertical distance obtained by the
projectile, the horizontal axis indicates measures
of time.
The average rate of change over a time interval is just the change in height over the
interval divided by the change in time. For the time interval [0, 2], the ball travels
from (0, 6) to (2, 70):
change in height
70 - 6 = 32
average rate of change = __ = _
2-0
change in time
The average rate of change is 32 feet per second. By symmetry, the time interval on
the way back down to the original height is [2, 4], and the average rate of change
will be the same, but with the opposite sign, or -32 feet per second.
B
A pumpkin emerges from an air cannon at a height of 24 feet above the ground with an
initial upward velocity of 192 feet per second.
a. How long does it take the pumpkin to reach its maximum height? What is that
height?
Use the model h(t) = -16t 2 + v 0t + h 0 with v 0 = 192 and h 0 = 24 :
h(t) = -16t 2 + 192 t + 24
192
b = -_ =
Time at maximum height: t = -_
6
2a
2(-16)
Substitute t = 6 to find the maximum height:
2
h(6) = -16(6) + 192(6) + 24 = 600
The pumpkin reaches its maximum height of 600 feet 6 seconds after it is fired.
b. What is the average rate of change in the pumpkin’s height over the time interval
from when it is fired to when it reaches its maximum height?
© Houghton Mifflin Harcourt Publishing Company
(
The average rate of change is 96 feet per second.
Module 3
A2_MTXESE353930_U2M03L1.indd 125
125
Lesson 3.1
)
⎡
⎤
For the time interval ⎢0, 6 ⎥, the ball travels from (0, 24) to 6, 600 :
⎣
⎦
600 - 24
change
in height
average rate of change = __ = __ = 96
change in time
6 -0
125
Lesson 1
2/22/14 5:09 AM
Reflect
9.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Technology
Complete the table for the model in Example 3 part B. What do you notice about the average rate of
change of the height over the intervals?
Time Interval
Height Over Time Interval
Average Rate of Change of Height Over Interval
from 0 s to 2 s
from 24 ft to 344 ft
from 2 s to 4 s
from 344 ft to 536 ft
96 ft/s
from 4 s to 6 s
from 536 ft to 600 ft
32 ft/s
from 6 s to 8 s
from 600 ft to 536 ft
-32 ft/s
from 8 s to 10 s
from 536 ft to 344 ft
-96 ft/s
from 10 s to 12 s
from 344 ft to 24 ft
Discuss with students the characteristics of the graph
of a function that models the motion of a projectile.
Have them graph a function from the lesson on a
graphing calculator and see how the attributes of the
graph reflect the real-world situation.
160 ft/s
-160 ft/s
The absolute value of the average rate of change of the height is the same for corresponding
intervals on either side of the axis of symmetry of the parabola.
10. You can rewrite the models in Example 3 parts A and B as h(t) = -16(t - 2) + 70 and
2
h(t) = -16(t - 6) + 600. Because the parameter a is the same for both models, -16, the graphs have the
same shape: they are both translations of -16x 2. How do the average rates of change you found reflect this
fact?
2
The average rate of change over the 2 second interval before the maximum is reached is the
same in absolute value as that over the 2 second interval after the maximum is reached.
11. The rate of change for a linear function, which is its slope, is constant. Use the table from the previous
question to describe the behavior of the average rate of change for a quadratic function.
The average rate of change is smaller in absolute value near the vertex of the graph, and
gets larger and larger in absolute value the farther you get from the vertex.
_
corresponds to time t in the model, the time does not depend on the initial height.
Your Turn
Use the projectile motion model to answer the questions.
13. A pebble is tossed from a cliff. It is released from a height 110 feet above the base of the cliff with an initial
upward velocity of 32 feet per second. How long does it take the pebble to reach its maximum height above
the base of the cliff? What is that height?
h(t) = -16t 2 + v 0t + h 0 = -16t 2 + 32t + 110;
32
b
=1
=Time at maximum height: t = 2a
2(-16)
2
h(1) = -16(1) + 32(1) + 110 = 126; max height: 126 ft after 1 second
_
Module 3
A2_MTXESE353930_U2M03L1 126
© Houghton Mifflin Harcourt Publishing Company
12. On what real-world parameters does the time it takes a projectile to reach its maximum height depend?
Explain your reasoning.
It depends on the parameter that gives the acceleration of gravity, –16 ft/s2, and the initial
b
velocity, but not on the initial height. This is because in x = - , the a corresponds to the
2a
“–16” in the model, and the b corresponds to the initial velocity in the model. Because x
_
126
Lesson 1
1/12/15 4:22 PM
Quadratic Functions in Vertex Form 126
Elaborate
ELABORATE
14. How is the vertex form of a quadratic function related to transformations of the graph of the parent
quadratic function ƒ(x) = x 2?
2
In the vertex form f(x) = a(x - h) + k, the h indicates the horizontal translation and k
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Math Connections
indicates the vertical translation. The value of a indicates whether the graph is a vertical
stretch or compression of the parent graph. The sign of a indicates whether the graph
opens upward (a > 0), or is a reflection across the line y = k (a < 0).
Discuss how transformations and the vertex form of a
quadratic function can be used to help graph the
function. Students should make a connection
between vertex form and the transformation of
f(x) = x 2. They should describe how points of the
graph can be determined, including the use of the
axis of symmetry to find reflections of points that lie
on the graph.
15. Essential Question Check-In What does the value of a in the vertex form of a quadratic function
reveal about the function?
The value of a reveals whether the graph opens up (a > 0) or down (a < 0), and whether
the graph is vertically stretched (|a| > 1) or compressed (|a| < 1).
Evaluate: Homework and Practice
1.
Why is the equation of the axis of symmetry
x = h? Because the axis of symmetry is a
vertical line that passes through the vertex of the
parabola, and the x-coordinate of the vertex is h.
When analyzing a quadratic function, what is
the advantage of writing the function in vertex
form? When the function is written in vertex form,
it is easier to determine its attributes, such as
direction of the opening of the graph, coordinates
of the vertex, maximum or minimum value, and
range. This also helps you find where the function is
increasing and decreasing.
Identify the axis of symmetry. Then give the function’s maximum or minimum value, domain,
range, and the intervals where the function is increasing or decreasing.
2.
© Houghton Mifflin Harcourt Publishing Company
SUMMARIZE THE LESSON
2
f(h - 5) = f(h + 5) = -19; f(h + 2) = f(h - 2) = -4; negative; the values
of f(x) closer to the axis of symmetry, x = h, are greater than the values of
f(x) farther from the axis of symmetry, so the graph opens downward.
QUESTIONING STRATEGIES
Is it possible for the equation of the axis of
symmetry to be x = 0? If so, under what
conditions? Yes. This would occur whenever the
vertex of the parabola is a point on the y-axis.
ƒ(x) = 5(x - 3) + 6
2
Lesson 3.1
3.
2
a = -7, h = 8, and k = 1
Axis of symmetry: x = 3
Axis of symmetry: x = 8
Minimum value of 6
⎧
⎧
⎫
⎫
D: ⎨x| x ∈ℜ⎬; R: ⎨y| y ≥ 6⎬
⎩
⎩
⎭
⎭
Decreasing on (-∞, 3)
Maximum value of 1
⎧
⎧
⎫
⎫
D: ⎨x| x ∈ℜ⎬; R: ⎨y| y ≤ 1⎬
⎩
⎩
⎭
⎭
Increasing on (-∞, 8)
Increasing on (3, +∞)
4.
ƒ(x) = -7(x - 8) + 1
a = 5, h = 3, and k = 6
Decreasing on (8, +∞)
ƒ(x) = -6(x + 4) - 8
2
5.
ƒ(x) = 2(x + 2) + 9
2
a = -6, h = -4, and k = -8
a = 2, h = -2, and k = 9
Axis of symmetry: x = -4
Axis of symmetry: x = -2
Maximum value of -8
⎧
⎧
⎫
⎫
D: ⎨x| x ∈ℜ⎬; R: ⎨y| y ≤ -8⎬
⎩
⎩
⎭
⎭
Increasing on (-∞, -4)
Minimum value of 9
⎧
⎧
⎫
⎫
D: ⎨x| x ∈ℜ⎬; R: ⎨y| y ≥ 9⎬
⎩
⎩
⎭
⎭
Decreasing on (-∞, -2)
Increasing on (-2, +∞)
Decreasing on (-4, +∞)
Module 3
A2_MTXESE353930_U2M03L1.indd 127
127
• Online Homework
• Hints and Help
• Extra Practice
For the function ƒ(x) = a(x - h) + k, ƒ(h + 5) = -19 and ƒ(h - 2) = -4.
Find ƒ(h - 5) and ƒ(h + 2). Is a positive or negative? Explain.
127
Lesson 1
2/22/14 5:14 AM
6.
EVALUATE
Match each quadratic function in vertex form with the coordinates of the graph of its vertex.
2
D
-h, k
A. ƒ(x) = a(x - h) + k
(
)
B. ƒ(x) = a(x - h) - k
A
(h, k)
C. ƒ(x) = a(x + h) - k
B
(h, -k)
D. ƒ(x) = a(x + h) + k
C
(-h, -k)
2
2
2
Write the quadratic function in vertex form by completing the square. Then identify
the maximum or minimum value and the value of x at which it occurs.
7.
ƒ(x) = x 2 + 6x - 3
(
f(x) = x 2 + 6x +
(_62 ) ) - 3 - (_26 )
2
8.
2
-10
-10
+ 11 - (_)
(_
2 ))
2
2
f(x) = (x - 5) - 14
2
2
f(x) = (x - (-3)) + (-12), so a = 1,
f(x) = (x - 5) + (-14), so a = 1,
2
2
h = -3, and k = -12.
h = 5, and k = -14.
The function f(x) has a minimum value
The function f(x) has a minimum value
of -12 when x = -3.
of -14 when x = 5.
ƒ(x) = 3x 2 + 24x + 53
(
f(x) = 3 x 2 + 8x +
(_8 ) ) + 53 - 3 (_8 )
2
2
10. ƒ(x) = 9x 2 + 18x - 3
(
2
f(x) = 9 x 2 + 2x +
2
f(x) = 3(x + 8x + 16) + 5
2
2
f(x) = (x 2 - 10x + 25) - 14
f(x) = (x + 3) - 12
9.
(
f(x) = x 2 - 10x +
f(x) = (x 2 + 6x + 9) - 12
ASSIGNMENT GUIDE
ƒ(x) = x 2 - 10x + 11
(_2 ) ) - 3 - 9 (_2 )
2
2
2
2
f(x) = 9(x 2 + 2x + 1) - 12
f(x) = 3(x + 4) + 5, so a = 3, h = -4,
f(x) = 9(x + 1) - 12
2
2
h = -1, and k = -12.
The function f(x) has a minimum value
of 5 when x = -4.
The function f(x) has a minimum value
of -12 when x = -1.
Module 3
Exercise
Depth of Knowledge (D.O.K.)
Mathematical Processes
1 Recall of Information
1.C Select tools
2–5
2 Skills/Concepts
1.B Problem solving model
6
2 Skills/Concepts
1.C Select tools
7–10
2 Skills/Concepts
1.B Problem solving model
11
2 Skills/Concepts
1.A Everyday life
12–14
2 Skills/Concepts
1.F Analyze relationships
1
Explore
Investigating Attributes of a
Quadratic Function in Vertex Form
Exercise 1
Example 1
Identifying the Attributes of a
Quadratic Function in Vertex Form
Exercises 2–6
Example 2
Converting from Standard Form to
Vertex Form
Exercises 7–10
Example 3
Analyzing a Projectile Motion Model
Exercises 11–14
When you see an equation of a quadratic
function in vertex form, how do you know if
the vertex is a minimum or maximum point? If a > 0,
the vertex is a minimum point, and if a < 0, the
vertex is a maximum point.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Critical Thinking
Ask students to consider the significance of h in
finding values of the domain for which a quadratic
function is increasing and for which it is decreasing.
Ask them how the value of a is useful for refining this
information.
Lesson 1
128
A2_MTXESE353930_U2M03L1.indd 128
© Houghton Mifflin Harcourt Publishing Company
f(x) = 9(x - (-1)) + (-12), so a = 9,
Practice
QUESTIONING STRATEGIES
2
and k = 5.
Concepts and Skills
1/10/15 4:37 PM
Quadratic Functions in Vertex Form 128
11. Multi-Step The height h above the roadway of the main cable
of the Golden Gate Bridge can be modeled by the function
7
1
h(x) = ____
x 2 - __
x + 500, where x is the distance in feet from the
9000
15
left tower.
VISUAL CUES
Although it is not necessary to graph the functions in
order to identify the attributes, some students may
find it helpful to draw a quick sketch.
x
h
a. Complete the square to write the function in vertex form.
-4200
-4200
1
1
_
x - 4200x + (_) ) + 500 - _(_)
9000 (
2
9000
2
1
h(x) = _(x - 4200x + 2100 ) + 500 - 490
9000
1
h(x) = _(x - 2100) + 10
h(x) =
2
2
2
2
AVOID COMMON ERRORS
2
9000
Students need to be careful to avoid making sign
errors when completing the square. Point out that
when the rule representing vertex form is simplified,
the result should be the original rule written in
standard form. Students can use this fact to perform a
quick check of the reasonableness of their results, and
perhaps catch any sign errors they may have made.
b. What is the vertex, and what does it represent?
The vertex is (2100, 10). The vertex represents that it is 2100 feet horizontally from the left
tower to the point at which the main cable reaches its lowest point of 10 feet above the
roadway
c. The left and right towers have the same height. What is the distance in feet between them? Explain.
The towers are 4200 feet apart because the axis of symmetry, x = 2100, means that each
tower is 2100 feet away from the lowest point in the main cable.
Use the projectile motion model to answer the questions.
12. A technician is launching an aerial firework from a 200-foot tower.
The firework’s upward velocity at launch will be 176 feet per second.
Professional fireworks are usually timed to explode as they reach their
highest point.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: Getty
Images Royalty Free
a. How long does it take the firework to reach its maximum height?
What is that height? Check your answer by graphing the model using a
graphing calculator.
h(t) = -16t 2 + v 0t + h 0 = -16t 2 + 176t + 200
176
b
== 5.5
Time at maximum height: t = 2a
2(-16)
2
h(5.5) = -16(5.5) + 176(5.5) + 200 = 684
_
The firework reaches its maximum height of 684 feet
5.5 seconds after it is launched.
Module 3
Exercise
A2_MTXESE353930_U2M03L1 129
129
Lesson 3.1
_
Lesson 1
129
Depth of Knowledge (D.O.K.)
Mathematical Processes
15–16
2 Skills/Concepts
1.C Select tools
17–18
2 Skills/Concepts
1.F Analyze relationships
1/12/15 6:41 PM
b. What is the average rate of change in the firework’s height
over first half second it is rising? over the last half second it is
rising?
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Technology
At t = 0, h = 200.
At t = 0.5, h(0.5) = -16(0.5) + 176(0.5) + 200 = 284.
change in height
284 - 200
84
=
=
= 168
average rate of change =
0.5 - 0
0.5
change in time
At t = 5.5, h = 684.
2
Students can use a graphing calculator to further
explore functions that model projectile motion. Have
them experiment with functions in both standard
and vertex form to see how changing the parameters
of the function alters the flight of the projectile.
Encourage students to share their observations with
the class.
__ _ _
At t = 5, h(5) = -16(5) + 176(5) + 200 = 680.
change in height
684 - 680
4
=
=
=8
average rate of change =
5.5 - 5
0.5
change in time
2
__ _ _
The average rate of change over the first half second of rise is 168 feet per second, and
over the last half second of rise is 8 feet per second.
13. A circus juggler on a high wire tosses a juggling pin upward with an initial vertical
velocity of 24 feet per second from a height of 32 feet.
a. How long does it take the pin to reach its maximum height?
What is that height?
h(t) = -16t 2 + v 0t + h 0 = -16t 2 + 24t + 32
b
= - 24 = 0.75
Time at maximum height: t = 2a
2(-16)
2
h(0.75) = -16(0.75) + 24(0.75) + 32 = 41
_
_
The pin reaches its maximum height of 41 feet 0.75 second after it is tossed.
b. The juggler misses the pin on its way down. How long after
the pin is tossed does it pass the juggler on the way down?
How do you know?
14. When height is measured in meters and time in seconds, the equation for projectile
motion becomes h(t) = -4.9t 2 + v 0t + h 0. A fishing boat with two people on it runs
out of gas while out in a bay. The people set off an emergency flare from a height of
2 meters above the water. The flare has an initial vertical velocity of 30 meters per
second. How long does it take the flare to reach its maximum height? What is that
height?
h(t) = -4.9t 2 + v 0t + h 0 = -4.9t 2 + 30t + 2
30
b
=≈ 3.06
Time at maximum height: t = 2a
2(-4.9)
2
h(3.06) = -4.9(3.06) + 30(3.06) + 2 ≈ 47.9
_
_
© Houghton Mifflin Harcourt Publishing Company
1.5 seconds; because the graph is symmetric, and the pin reaches its maximum height
0.75 seconds after being released, it will be at that same height again on its way down
0.75 seconds later, and 0.75s + 0.75s = 1.5s.
The flare reaches its maximum height of about 48 meters just over 3 seconds after being
fired.
Module 3
A2_MTXESE353930_U2M03L1.indd 130
130
Lesson 1
2/20/14 12:13 AM
Quadratic Functions in Vertex Form 130
CONNECT VOCABULARY
H.O.T. Focus on Higher Order Thinking
15. Explain the Error Two attempts to write ƒ(x) = 2x 2 - 8x in vertex form are shown.
Which is incorrect? Explain the error.
Connect the use of standard form for quadratic
functions to the standard form of an equation, as
opposed to its vertex form, or to the standard form of
a number, as opposed to its expanded form. Remind
students that sometimes we rewrite standard forms of
equations, numbers, or functions in other forms to
facilitate the extraction of information or for solving
purposes. In this case, the vertex form of a quadratic
function yields information about the direction of the
opening of the graph, the vertex, maximum and
minimum values, and so on.
A. ƒ(x) = 2x 2 - 8x
B. ƒ(x) = 2x 2 - 8x
ƒ(x) = 2(x 2 - 4x)
ƒ(x) = 2(x 2 - 4x)
ƒ(x) = 2(x 2 - 4x + 4) -4
ƒ(x) = 2(x 2 - 4x + 4) -8
2
ƒ(x) = 2(x - 2) - 4
2
ƒ(x) = 2(x - 2) - 8
Part A is incorrect. In the third step, the number 4 is added inside the
parentheses. Since the expression in parentheses is multiplied by 2, the
total number added to the function expression is 8. So, the number 8,
not 4, must be subtracted from the expression to keep it equivalent.
16. Communicate Mathematical Ideas You can rewrite a quadratic function that is
in standard form in vertex form by completing the square. How can you also use the
fact that the maximum or minimum of a quadratic function in standard form occurs
b
when x = -__
to rewrite it in vertex form?
2a
PEER-TO-PEER DISCUSSION
In vertex form, h gives the x-value of the minimum or maximum, so
Ask students to discuss with a partner why the
motion of a projectile cannot be modeled by a linear
function. Possible answer: The model gives height
as a function of time, and the height of the
projectile increases and then decreases. Thus, a
linear model is not possible, since linear functions
either increase or decrease, or remain constant.
b
function value at -___
. The parameter a is the coefficient of x 2 in both
2a
Have students describe how to write a quadratic
function in vertex form, and how to use the result to
describe the attributes of the function.
( ( )) + f(-___).
b
forms. So, f(x) = ax 2 + bx + c becomes f(x) = a x - -___
2a
g(x) = (x + 1) - 3 = x 2 + 2x + 1 - 3 = x 2 + 2x - 2.
2
18. Justify Reasoning Over a given time interval, a golf ball in flight has an average
rate of change in height of 0. Does this mean the ball was moving horizontally?
Explain.
No; A quadratic function is always increasing or always decreasing on
each side of the vertex. The interval had to be symmetric about the
vertex, so the ball increased in height and then decreased in height by the
same amount, which made the average zero.
A2_MTXESE353930_U2M03L1.indd 131
Lesson 3.1
b
2a
Yes; you can complete the square for f(x) to show that the expressions are
equivalent, or, more simply, expand the binomial in g(x) and simplify:
Module 3
131
2
17. Analyze Relationships The functions ƒ and g are defined by ƒ(x) = x 2 + 2x - 2
2
and g(x) = (x + 1) - 3. Do f and g represent the same function? Explain.
© Houghton Mifflin Harcourt Publishing Company
JOURNAL
b
. Also, the maximum or minimum is k. This corresponds to the
h = -___
2a
131
Lesson 1
1/10/15 4:37 PM
Lesson Performance Task
AVOID COMMON ERRORS
Students may assume that since only the linear factor
is changing in part d., they can add (32 - 20)x to the
function value and use the same x-coordinate of the
41 , to find the difference in maximum
vertex, x = ___
4
height. Remind them that a change in the coefficient
of x also changes the x-value of the vertex, and the
new maximum height is dependent on this value, so
this will not result in a correct answer.
A player bumps a volleyball from a height of 4 feet with an initial vertical velocity of 20 ft/s.
a. Use the function h(t) = -16t 2 + v 0t + h 0 , where v 0 is the
initial velocity and h 0 is the initial height to write a function h
in standard form for the ball’s height in feet in terms of time t
in seconds after the ball is hit.
b. Complete the square to rewrite h in vertex form.
c. What is the maximum height of the ball?
d. Suppose the volleyball were hit under the same conditions, but
with an initial velocity of 32 ft/s. How much higher would the
ball go?
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Technology
a. The function would be h(t) = -16t 2 + 20t + 4.
h(t) = -16t 2 + 20t + 4
5
h(t) = -16 t 2 - t + 4
4
25
5
25
2
+ 4 + 16
h(t) = -16 t - t +
4
64
64
2
5
25
+4+
h(t) = -16 t 4
8
2
5
41
h(t) = -16 t +
4
8
41
1
c. The maximum height will occur at
or 10 feet.
4
4
d. h(t) = -16t 2 + 32t + 4
b.
(
(
(
(
_)
_ _)
_)
_
_) _
(_)
_
A spreadsheet can be used to do some of the basic
calculations in order to reduce the possibility of
errors. Keep in mind that it will produce decimal
results rather than fractions.
_
h(t) = -16(t - 2t + 1) + 4 + 16(1)
2
h(t) = -16(t - 1) + 4 + 16
2
h(t) = -16(t - 1) + 20
2
If the initial velocity were 32 ft/s then the ball would go 9
Module 3
132
QUESTIONING STRATEGIES
© Houghton Mifflin Harcourt Publishing Company • ©Adam Brown
Photography/Getty Images
h(t) = -16(t 2 - 2t) + 4
_3 feet higher.
4
If the ball is hit with initial velocity 32 ft/s
instead of 20 ft/s, how much longer would it
take for the ball to reach its maximum height?
3 second longer.
It would take __
8
How is this difference in time represented on a
graph of the functions? It is the difference
between the x-coordinates of the vertices.
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Lesson 1
EXTENSION ACTIVITY
A2_MTXESE353930_U2M03L1.indd 132
Have students consider the greatest height to which a ball can be hit in their
indoor gymnasium without encountering the ceiling or any lights, scoreboards, or
other obstructions. They can then try to find the value of v0 that will produce this
height.
1/10/15 4:37 PM
Students might also research the rules concerning a ball that does hit something
above the court; these rules vary in stating whether the ball is dead or playable.
Quadratic Functions in Vertex Form 132