LESSON
3.2
Name
Writing Quadratic
Functions
Class
3.2
A2.4.B: Write the equation of a parabola using given attributes, including vertex, …axis of
symmetry, and direction of opening.
Explore
The student is expected to:
A2.4.B

Mathematical Processes
A2.1.F
Identifying Information for Writing a Quadratic
Function in Vertex Form
The parabola opens downward. What do you know about the value of a in the equation of
the parabola? Insert <, >, or =.
a < 0
The student is expected to analyze mathematical relationships to connect
and communicate mathematical ideas.

The axis of symmetry is x = 1. What does this tell you about the vertex of the parabola?
The x -value of the vertex of the parabola is 1 .
Language Objective
2.H.2, 2.I.1, 2.I.3, 3.D.1, 3.D.2, 5.G.2

Students work in pairs or small groups to speak and listen to and give oral
clues about graphs of quadratic functions.
The range of the parabola is (-∞, 4). What does this tell you about the vertex of the
parabola?
The y -value of the vertex of the parabola is 4 .
Reflect
ENGAGE
PREVIEW: LESSON
PERFORMANCE TASK
Resource
Locker
A given parabola has an axis of symmetry of x = 1, a range of -∞ to 4, and opens downward. What does this
information tell you that you can use to write the equation of this quadratic function in vertex form?
Write the equation of a parabola using given attributes, including vertex,
focus, directrix, axis of symmetry, and direction of opening.
1.
© Houghton Mifflin Harcourt Publishing Company
Possible answer: The information includes the
direction of opening, axis of symmetry, maximum/
minimum values, and vertex. Also, consider how
knowledge of one or more (non-vertex) points and
some other information may allow you to solve for
unknown parameters.
Writing Quadratic Functions
Essential Question: What information identifies the parameters for writing a quadratic
function in vertex form?
Texas Math Standards
Essential Question: What information
identifies the parameters for writing a
quadratic function in vertex form?
Date
What other value do you need to write the equation of this parabola in vertex form? Explain what you
would need to find this value and how you would find it.
The value of a; You would need another point on the parabola. You would substitute the
x- and y-values of that point onto the equation f(x) = a(x - 1) + 4 and solve for a.
2
2.
Suppose the range was (4, ∞) and the parabola opened upward. What would be different about the
equation of the parabola? What would be the same?
2
The equation would still be f(x) = a(x - 1) + 4, but the value of a would be positive.
Module 3
ges
EDIT--Chan
DO NOT Key=TX-A
Correction
Lesson 2
133
gh "File info"
made throu
must be
Date
nctions
adratic Fu
Name
Writing Qu
quadratic
writing a
eters for
the param
Resource
identifies
Locker
information
of
form?
ion: What
vertex, …axis
in vertex
including
function
attributes,
la using given
ic
on of a parabo
a Quadrat
the equati
g.
A2.4.B: Write direction of openin
for Writing
and
symmetry,
Information
ing
this
tify
does
Form
What
Iden
downward.
in Vertex
Explore
and opens
form?
Function
-∞ to 4,
n in vertex
of
functio
range
= 1, a
quadratic
etry of x
on of this
axis of symm write the equati
on of
la has an
use to
in the equati
A given parabo you that you can
value of a
tell
about the
information
3.2
Quest
Essential
A2_MTXESE353930_U2M03L2 133
do
ard. What
la opens downw or =.
The parabo
<, >,
la? Insert
the parabo

a < 0
The axis

of symmetry
-value of
The x
The range
parabola?

What does
.
la is 1
of the parabo
is x = 1.
the vertex
of
of the vertex
HARDCOVER PAGES 99106
Turn to these pages to
find this lesson in the
hardcover student
edition.
you know
about
this tell you
la is (-∞,
of the parabo
-value
The y
4). What
does this
the vertex
tell you about
la?
of the parabo
the vertex
of the
.
la is 4
the parabo
n what you
form? Explai
la in vertex
of this parabo
itute the
equation
it.
would subst
to write the
would find
ola. You
you need
a.
how you
value do
the parab
value and
solve for
What other
2
point on
to find this
another
1) + 4 and
would need
= a(x would need
tion f(x)
of a; You
the equa
The value
point onto
that
of
the
es
nt about
x- and y-valu
be differe
d. What would
upwar
d
ive.
la opene
d be posit
the parabo
of a woul
(4, ∞) and
the value
be the same?
2
but
range was
4,
would
the
+
se
)
What
(x - 1
2. Suppo of the parabola?
f(x) = a
d still be
equation
tion woul
equa
The
Reflect
1.
© Houghto
n Mifflin
Harcour t
Publishin
y
g Compan
View the online Engage. Discuss the fact that
stopping distance for a vehicle such as a car or truck
can be modeled by a quadratic function of speed, and
describe the usefulness of such a function. Then
preview the Lesson Performance Task.
Class
Lesson 2
133
Module 3
133
Lesson 3.2
L2 133
0_U2M03
SE35393
A2_MTXE
20/02/14
5:21 AM
20/02/14 5:21 AM
Explain 1
Writing a Quadratic Function Given a Parabola’s
Vertex, Direction of Opening, and Another Point
EXPLORE
Recall that the vertex form of a quadratic function is ƒ(x) = a(x - h) + k, where (h, k) is the vertex of the function.
Also, recall that if a < 0, then the parabolic graph of the function opens downward and the vertex is the maximum
point on the parabola, and that if a > 0, then the parabolic graph of the function opens upward and the vertex is the
minimum point on the parabola. Knowing this information, and one other point on the parabola, will allow you to
write the equation of the quadratic function associated with a given parabola.
2
Example 1

Identifying Information for Writing a
Quadratic Function in Vertex Form
Use the vertex, direction of opening, and another point on each parabola
to write the equation of the quadratic function in vertex form.
QUESTIONING STRATEGIES
vertex: (3, 2); point: (5, -6); direction of opening: downward
Can you use any two points on a quadratic
function graph to write its equation?
Explain. No; many different quadratic equations
can fit two points.
h = 3, k = 2
Identify h and k from the vertex.
Substitute h and k into the vertex form of a quadratic function.
ƒ(x) = a(x - 3) + 2
Substitute the other point into this equation to solve for a.
-6 = a(5 - 3) + 2
2
2
-6 = a(2) + 2
2
-6 = 4a + 2
EXPLAIN 1
-8 = 4a
-2 = a

Writing a Quadratic Function Given
a Parabola’s Vertex, Direction of
Opening, and Another Point
ƒ(x) = -2 (x - 3) + 2
Substitute a into the vertex form of the function.
2
vertex: (-1, 5); point: (0, 8); direction of opening: upward
h = -1 , k =
Identify h and k from the vertex.
Substitute h and k from the vertex form of a quadratic function.
Substitute the other point into this equation to solve for a.
5
(
)+ 5
= a( 0 + 1 ) + 5
= a( 1 ) + 5
ƒ(x) = a x - -1
8
2
8 = 1 a+ 5
3 = 1 a
3 =a
Substitute a into the vertex form of the function.
Module 3
ƒ(x) = 3
134
Students sometimes make an error when substituting
for h in the vertex form of the function. Help them
avoid this error by pointing out that h must be the
number subtracted from x. Illustrate how to handle
both positive and negative values of h.
2
(
x+1
)
2
+ 5
© Houghton Mifflin Harcourt Publishing Company
8
AVOID COMMON ERRORS
2
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Reasoning
Discuss with students how, no matter what point on
the parabola is chosen to substitute for x and y, the
value of a will be the same. Lead them to recognize
that this is because every value of y in the range of the
function is related to its corresponding value of x in
the same way.
Lesson 2
PROFESSIONAL DEVELOPMENT
A2_MTXESE353930_U2M03L2.indd 134
Integrate Mathematical Processes
This lesson provides an opportunity to address Mathematical Process
TEKS A2.1.F, which calls for students to “analyze mathematical relationships to
connect and communicate mathematical ideas.” Students analyze information
about the graph of a quadratic function, and use the information to write the rule
for the corresponding function. They learn how different sets of information
require different approaches, and how these approaches can be used to write
quadratic functions that model real-world situations.
1/10/15 5:11 PM
QUESTIONING STRATEGIES
How can you check that you’ve written the
correct function? You can choose another
point from the graph and show that its coordinates
satisfy the equation. You can also use a graphing
calculator to check that your function produces the
same graph.
Writing Quadratic Functions
134
Your Turn
EXPLAIN 2
3.
vertex: (3, -3); point: (2, -7); direction of opening: downward
h = 3, k = -3
Writing a Quadratic Function Given
a Parabola’s Axis of Symmetry,
Direction of Opening, and Two
Non-vertex Points
f(x) = a(x - 3) - 3
2
-7 = a(2 - 3) - 3
2
-7 = a(-1) - 3
2
-7 = a - 3
-4 = a
f(x) = -4(x - 3) - 3
QUESTIONING STRATEGIES
2
Writing a Quadratic Function Given a
Parabola’s Axis of Symmetry, Direction of
Opening, and Two Non-Vertex Points
Explain 2
Why do you need to write two equations to
find the rule for the function? Substituting
for h, x, and y creates an equation in two variables.
To solve for two variables, two equations are
needed.
When given the axis of symmetry of a parabola, two non-vertex points, and the direction of opening,
you can write the quadratic function in vertex form. Recall that the axis of symmetry of a parabola
always passes through the vertex. Therefore, if the line of symmetry is x = h, then the x-value of the
vertex is h. You can use the two non-vertex points to set up a system of two linear equations in a and k
and then solve those equations for a and k. The direction the parabola opens determines whether the
value of a will be positive or negative.
Could there be more than one quadratic
function that contains the given points? Yes,
but only one that also has the given axis of
symmetry.
Example 2

Write the equation of the quadratic function in vertex form for the parabola
that will satisfy the given conditions.
The parabola opens upward, has x = 4, as the axis of symmetry, and contains the non-vertex
points (3, 5) and (7, 12).
The axis of symmetry, x = 4, gives the value of h in the vertex form of the function.
ƒ(x) = a(x - 4) + k
© Houghton Mifflin Harcourt Publishing Company
2
Substitute each of the two points for x and ƒ(x) into the function to obtain a system of two
linear equations in two unknowns.
ƒ(x) = a(x - 4) + k
2
5 = a(3 - 4) + k
ƒ(x) = a(x - 4) + k
2
12 = a(7 - 4) + k
2
5 = a(-1) + k
2
5=a+k
⎧ a+k=5
The system is ⎨
a+k=5
2
Substitute (3, 5).
12 = a(3) + k
2
Simplify.
Substitute (7, 12).
Simplify.
12 = 9a + k
. Solve the first equation for k.
⎩ 9a + k = 12
k=5-a
Module 3
135
Lesson 2
COLLABORATIVE LEARNING
A2_MTXESE353930_U2M03L2.indd 135
Small Group Activity
Have students work in groups of 3-4. Each group brainstorms to identify a
real-world situation that can be modeled by a quadratic function, then researches
to find data to use in posing a problem involving identification of the function
(similar to Example 3). They then create a poster illustrating the scenario, stating
the problem and its solution and give an example of how the function can be used
to answer questions about the scenario.
135
Lesson 3.2
14/01/15 8:12 PM
Substitute for k in the second equation and
solve for a.
Substitute a = _78 into the first equation and
solve for k.
12 = 9a + k
a+k=5
7 +k=5
_
8
k = 5 - _78
33
k=_
8
12 = 9a + 5 - a
Substitute.
12 = 8a + 5
Simplify.
7 = 8a
Subtract 5 from each side.
Substitute.
Solve for k.
Simplify.
7
a=_
Divide each side by 8.
8
2
33
Therefore, the equation in vertex form is ƒ(x) = _78 (x - 4) + __
.
8
B
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Critical Thinking
Challenge students to consider how they might write
the rule for a quadratic function given three
non-vertex points. Lead them to see that three
equations would be needed, as it would be necessary
to solve for three variables, a, h, and k.
The parabola opens upward, has x = -5 as the axis of symmetry, and contains the
non-vertex points (-8, 7) and (5, 20).
The axis of symmetry, x = -5 gives the value of h in the vertex form of the function.
(
) +k
2
ƒ(x) = a x - (-5)
Substitute each of the two points for x and f(x) into the function to obtain a system of
two linear equations in two unknowns.
ƒ(x) = a(x + 5) + k
ƒ(x) = a(x + 5) + k
2
7 = a(-8 + 5) + k
2
7 = a(-3) + k
2
2
( 5 + 5) + k
20 = a ( 10 ) + k
Substitute (-8, 7) .
2
20 = a
2
Simplify.
7 = 9a + k
Substitute (5, 20).
Simplify.
20 = 100a + k
⎧ 9a + k = 7
. Solve the first equation for k.
The system is ⎨
⎩ 100a + k = 20
9a + k = 7
k = 7 - 9a
Substitute a =
Substitute for k in the second equation and
solve for a.
(
9a + k = 7
1
_
7
20 = 100a + 7 - 9a
Substitute.
20 = 91a + 7
Simplify.
13 = 91a
Substract 7 from each side.
13
a=_
91
1
a= _
7
9
)+k=7
Therefore, the equation in vertex form is
(
k=7-9
k=7-
40
k= _
7
Divide each side by 91.
Simplify.
into the first equation.
f (x) =
9
_
7
1
_
7
)
© Houghton Mifflin Harcourt Publishing Company
20 = 100a + k
1
_
7
Substitute.
Solve for k.
Simplify.
Substract.
40
_1 (x + 5) + _
2
7
7
.
Reflect
4.
Given the axis of symmetry and two points, what else can you immediately determine about the parabola?
The direction in which the parabola opens and, using symmetry, the values of two more
points can be determined.
Module 3
136
Lesson 2
DIFFERENTIATE INSTRUCTION
A2_MTXESE353930_U2M03L2.indd 136
Multiple Representations
1/10/15 5:11 PM
Some students may need to draw a sketch illustrating the given information before
they can assess how to proceed. By attempting to draw the parabola being
described, students attach meaning to the verbal descriptions and are better able
to identify the “missing pieces” of information. They then can determine what
needs to be done to find these missing pieces of information and to solve the
problem.
Writing Quadratic Functions
136
Your Turn
CONNECT VOCABULARY
Write the equation in vertex form for the parabola that will satisfy the given
conditions.
Relate the word quadratic, whose root means four or
square, to a square. Remind students that they can
solve a quadratic equation by completing the square,
and that the coordinate plane where we graph
quadratic functions is in quadrants.
5.
The parabola opens downward, has x = 6 as an axis of symmetry, and contains the non-vertex
points (15, 3) and (13, 6).
The axis of symmetry, x = 6, gives the value of h in f(x) = a(x - h) + k.
2
Substitute each of the two points in the function to obtain a system of two linear
equations.
3 = a(15 - 6) + k
2
3 = a(9) + k
2
Substitute (15, 3).
Simplify.
3 = 81a + k
6 = a(13 - 6) + k
2
6 = a(7) + k
2
Substitute (13, 6).
Simplify.
6 = 49a + k
⎧ 81a + k = 3
The system is ⎨
. Solve the first equation for k.
⎩ 49a + k = 6
81a + k = 3
k = 3 - 81a
Substitute for k in the second equation and solve for a.
6 = 49a + 3 - 81a
6 = -32a + 3
3 = -32a
Subtract 3 from each side.
3
a=Divide each side by -32.
32
3
Substitute a = into the first equation.
⎛ 3 ⎞ 32
⎟+k=3
81⎜Substitute.
⎝ 32 ⎠
⎛ 3 ⎞
⎟
Solve for k.
k = 3 - 81⎜⎝ 32 ⎠
243
Simplify.
k=3+
32
339
Add.
k=_
32
2
3
339
f(x) = -_ (x - 6) + _
32
32
_
_
© Houghton Mifflin Harcourt Publishing Company
Substitute.
Simplify.
_
_
_
Module 3
137
Lesson 2
LANGUAGE SUPPORT
A2_MTXESE353930_U2M03L2.indd 137
Connect Vocabulary
Have each student sketch a graph of a parabola on a card, and write a quadratic
function in any form on another card. Then have them write clues about the graph
and the function. Sample clues include: “The parabola opens upward/downward.
Its axis of symmetry is ____; the vertex is at point _______ . The function is
written in standard/vertex form. The function’s graph will open downward/
upward. The function shows x-intercepts at ____and ____.” Provide sentence
stems if needed to help students begin their clues. Have other students then
choose the card that fits the clue.
137
Lesson 3.2
1/10/15 5:11 PM
Modeling with Quadratic Functions
Explain 3
EXPLAIN 3
You can find the solution to a real-world problem using information given in the problem to write a quadratic
function and use the function to estimate an answer.

Modeling with Quadratic Functions
A bald eagle grabs a fish from a mountain lake and flies to
an altitude of 256 feet above the lake. At that point, the fish
manages to squirm free and falls back down into a river flowing
out of the lake. After falling 3 seconds, the fish is 112 feet above
the surface of the lake it was taken from. If the surface of the
river where the fish lands is 144 feet below the surface level of
the lake, for how many seconds did the fish fall?
AVOID COMMON ERRORS
Students sometimes make errors when writing
functions that model real-world situations because
they do not correctly identify the measures that are
represented by each of the variables. Encourage
students to read the problem, analyze the situation,
identify both the dependent and independent
measures, and assign them correctly to x and y.
Analyze Information
Identify the important information.
• The function modeling the height of
the fish is quadratic with respect to
• The vertex of the parabola is at
• A second point on the parabola is
time .
(0, 256) .
(3, 112) .
Formulate a Plan
You want to find the number of seconds the fish fell.
The vertex of the parabola is given, so write the function h(t) in vertex
2
form: h(t) = a(t - h) + k .
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Modeling
Use the non-vertex point to find a.
Then find t such that h(t) =
Solve
The vertex is (0, 256) . Substitute these values into h(t).
h(t) = a(t - 0) + 256
2
Substitute a into the function.
Substitute the values of the non-vertex point
into the function and solve for a.
h(t) = a(t - 0) 2 + 256
112
112
= a
= a
(
(
3
3
-144 = 9a
a=
)
h(t) =
2
-144
- 0 + 256
)
2
25
= -16t + 256
A2_MTXESE353930_U2M03L2.indd 138
2
= t2
t 2 = 25
-16
t =
Module 3
t 2 + 256
2
-400 = -16t
+ 256
The fish hits the surface of the river
-16
5
Discuss with students how to correlate the real-world
information with the mathematical attributes of the
function that will be used to model the situation.
Some students may benefit from drawing a sketch
and incorporating the information from the problem
into the sketch.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Alan and
Sandy Carey/ Getty Images
-144 .
QUESTIONING STRATEGIES
What is the axis of symmetry in a real-world
situation that can be modeled by a quadratic
function? The axis of symmetry is an imaginary
vertical line that passes through the maximum or
minimum point reached by the object. It
corresponds to a value of the independent variable,
for example, “the time when the ball reaches its
maximum height.”
5
seconds after being dropped.
138
Lesson 2
2/22/14 5:35 AM
Writing Quadratic Functions
138
Justify And Evaluate
ELABORATE
5
Check your answer by substituting
into h(t) and seeing if
2
2
)
(
-16
5
0
+
256
= -16(5) + 256 = -400 + 256 = -144
-144
h(t) =
:
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Math Connections
Your Turn
6.
Discuss with students the connection between the
geometry of the graph of the function (that is, the
geometric nature of the parabola) and the algebraic
rule for the function. Help them to see how a
function that maps elements of the domain onto
elements of the range via a quadratic function
produces a graph with a parabolic shape.
During a competition, an archer fires an arrow at a target
on a hill. The arrow reaches a maximum height of 127 feet
2.75 seconds after it was fired. The arrow hits the target 5 seconds
after the archer releases it. If the arrow is released from a height of
6 feet, how far above the ground on which the archer is standing
is the target? Assume the relationship between time and height is
quadratic.
The vertex is (2.75, 127) and a second non-vertex point
is (0, 6).
The height of the arrow above the ground as a function of time is h(t) = a(t - 2.75) + 127.
2
Evaluate the function at the non-vertex point (0, 6) to find a.
6 = a(0 - 2.75) + 127
2
6 = a(-2.75) + 127
h(5) = -16(2.25) + 127
2
QUESTIONING STRATEGIES
How can you use information about the graph
of a quadratic function to write the rule for
the function in vertex form? You can see what
the information tells you about the values of a, h,
and/or k, and use points on the parabola to help
determine the values of any of these variables for
which no information is given. You can then
substitute the values into the equation
2
y = a(x - h) + k.
h(5) = -81 + 127
h(5) = 46
a = -16
h(t) = -16(t - 2.75) + 127
2
The target is 46 feet above the ground where the archer is standing.
Elaborate
7.
Discussion If you are given two points in the coordinate plane, can you create a parabola that passes
through both points and opens in a specified direction?
Yes. If the parabola opens upward, make the point with the lesser y-coordinate the vertex
and use the other point to find a in the vertex form of the function. If the parabola opens
downward, make the point with the greater y-coordinate the vertex and use the other
point to find a in the vertex form of the function.
If the y-coordinates of the points are equal, average the x-coordinates to find the line that
is the axis of symmetry. Then select a point on that line to be the vertex. If the parabola
opens upward, choose a point on the axis of symmetry with a y-coordinate less than the
given y-coordinates. If the parabola opens downward, pick a point on the axis of symmetry
with a y-coordinate greater than the given y-coordinates.
8.
Essential Question Check-in Describe the minimum amount of information required to create
a unique parabola. Explain your answer.
The minimum amount of information required is the vertex and a second, non-vertex,
point. Having just the zeros, or x-intercepts, will give a family of parabolas with those
zeros. Any other two points in the plane will result in at least two parabolas.
A2_MTXESE353930_U2M03L2.indd 139
Lesson 3.2
h(5) = -16(5.0625) + 127
-121 = 7.5625a
Module 3
139
2
6 = 7.5625a + 127
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Erik
Isakson/Tetra Images/Corbis
SUMMARIZE THE LESSON
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Alan and
Sandy Carey/ Getty Images
Is it important to know whether a parabola
opens up or down when solving for a?
Explain. No. The sign of a will be determined
algebraically.
Find h(5).
h(5) = -16(5 - 2.75) + 127
2
139
Lesson 2
1/10/15 5:11 PM
Evaluate: Homework and Practice
1.
A given parabola has an axis of
2.
symmetry of x = 5, a range of 7 to
∞, and opens upward. What does this
information tell you that you can use
to write the equation of this quadratic
function in vertex form?
a > 0 and the vertex is (5, 7).
EVALUATE
A given parabola has an axis of
symmetry of x = -4, a range of -∞
to 10, and opens downward. What
does this information tell you that you
can use to write the equation of this
quadratic function in vertex form?
• Online Homework
• Hints and Help
• Extra Practice
a < 0 and the vertex is (-4, 10).
ASSIGNMENT GUIDE
Use the vertex, direction of opening, and another point on each
parabola to write its equation in vertex form.
3.
vertex: (0, 0); point: (2, 8); opens upward
4.
h = 2 and k = 1, so f(x) = a(x - 2) + 1.
h = 0 and k = 0, so f(x) = a(x - 0) + 0.
2
2
Substitute the values for x and f(x)
and solve for a.
Substitute the values for x and f(x)
and solve for a.
8
8
8
2
=
=
=
=
a(2 - 0) + 0
2
a(2)
4a
a
2
2
2
2
1
= a(3 - 2) + 1
2
= a(1) + 1
=a+1
=a
2
f(x) = 1(x - 2) + 1, or f(x) = (x - 2) + 1
f(x) = 2(x - 0) + 0, or f(x) = 2x 2
5.
vertex: (2, 1); point: (3, 2); opens upward
2
2
2
vertex: (-6, 4); point: (-5, 0); opens downward 6.
vertex: (-5, 10); point: (-8, -8); opens downward
h = -6 and k = 4, so f(x) = a(x + 6) + 4.
h = -5 and k = 10, so f(x) = a(x + 5) + 10.
Substitute the values for x and f(x)
and solve for a.
Substitute the values for x and f(x)
and solve for a.
2
2
-8
-8
-8
-18
-2
2
-4 = a
f(x) = -4(x + 6) + 4
2
Module 3
=
=
=
=
=
a(-8 + 5) + 10
2
a(-3) + 10
2
9a + 10
9a
a
f(x) = -2(x + 5) + 10
2
Exercise
Depth of Knowledge (D.O.K.)
Mathematical Processes
1 Recall of Information
1.G Explain and justify arguments
2–10
2 Skills/Concepts
1.F Analyze relationships
11–14
2 Skills/Concepts
1.B Problem solving model
15–20
1 Recall of Information
1.G Explain and justify arguments
21
2 Skills/Concepts
1.F Analyze relationships
22–23
2 Skills/Concepts
1.F Analyze relationships
1
Practice
Explore
Identifying Information for Writing a
Quadratic Function in Vertex Form
Exercises 1–2,
15–20
Example 1
Writing a Quadratic Function Given
a Parabola’s Vertex, Direction of
Opening, and Another Point
Exercises 3–6
Example 2
Writing a Quadratic Function Given
a Parabola’s Axis of Symmetry,
Direction of Opening, and Two
Non-vertex Points
Exercises 7–10
Example 3
Modeling with Quadratic Functions
Exercises 11–14
QUESTIONING STRATEGIES
What do x and y represent in the equation
2
y = a(x - h) + k? the coordinates of any
point that lies on the graph of the function
Lesson 2
140
A2_MTXESE353930_U2M03L2.indd 140
© Houghton Mifflin Harcourt Publishing Company
0 = a(-5 + 6) + 4
2
0 = a(1) + 4
0 = a+4
Concepts and Skills
1/10/15 5:11 PM
Writing Quadratic Functions
140
Write the equation in vertex form of the parabola that has the given axis of symmetry,
points, and direction of opening.
VISUAL CUES
7.
Drawing a sketch of the given information will
provide a visual cue for considering the general
reasonableness of the equation students write. In
particular, a sketch may help students determine the
correct sign of a, as well as a reasonable
approximation of k, when these values are not given.
axis of symmetry: x = 3; points: (1, 9), (4, 3); direction of opening: upward
From the axis of symmetry, f(x) = a(x - 3) + k.
2
Substitute each of the two points into the function to obtain a system of
two linear equations.
2
9 = a(1 - 3) + k
2
9 = a(-2) + k
9 = 4a + k
Substitute (1, 9).
Simplify.
3 = a(4 - 3) + k
Substitute (4, 3).
2
3 = a(1) + k
Simplify.
3 = a+k
⎧4a + k = 9
The system is ⎨
. Subtract the second equation from the first
⎩ a+k=3
equation, and then solve for a.
3a = 6
a=2
2
Substitute a = 2 into the second equation and solve for k.
2+k = 3
k= 1
f(x) = 2(x - 3) + 1
2
8.
axis of symmetry: x = -2; points: (-3, 2), (0, -13); direction of opening: downward
From the axis of symmetry, f(x) = a(x + 2) + k.
2
© Houghton Mifflin Harcourt Publishing Company
Substitute each of the two points into the function to obtain a system of
two linear equations.
2
2 = a(-3 + 2) + k
2
2 = a(-1) + k
2 = a+k
Substitute (-3, 2).
Simplify.
-13 = a(0 + 2) + k
Substitute (0, -13).
2
-13 = a(2) + k
Simplify.
-13 = 4a + k
⎧ a+k=2
The system is ⎨
. Subtract the first equation from the second
⎩ 4a + k = -13
equation, and then solve for a.
3a = -15
a = -5
2
Substitute a = -5 into the first equation and solve for k.
-5 + k = 2
k= 7
f(x) = -5(x + 2) + 7
2
Module 3
A2_MTXESE353930_U2M03L2.indd 141
141
Lesson 3.2
141
Lesson 2
1/10/15 5:11 PM
9.
axis of symmetry: x = -3; points: (-7, -2), (-5, -8); direction of opening: upward
From the axis of symmetry, f(x) = a(x + 3) + k.
CRITICAL THINKING
Substitute each of the two points into the function to obtain a system of
Prompt students to consider why it is not possible to
determine the equation of the function if they know
only the vertex and the axis of symmetry of the
parabola. Students should recognize that infinitely
many parabolas have the same vertex and axis of
symmetry, and that more information is needed to
determine the equation of the specific parabola.
2
two linear equations.
2
-2 = a(-7 + 3) + k
2
(
)
-2 = a -4 + k
-2 = 16a + k
Substitute (-7, -2).
Simplify.
-8 = a(-5 + 3) + k
Substitute (-5, -8).
2
-8 = a(-2) + k
Simplify.
-8 = 4a + k
⎧16a + k = -2
The system is ⎨
. Subtract the second equation from the first
⎩ 4a + k = -8
equation, and then solve for a.
2
12a = 6
1
a=_
2
1 into the first equation and solve for k.
Substitute a = _
2
⎛_
⎞
1
16⎜ ⎟ + k = -2
⎝2⎠
8 + k = -2
k = -10
1 (x + 3)2 - 10
f(x) = _
2
10. axis of symmetry: x = 4; points: (3, -2), (6, -26); direction of opening: downward
From the axis of symmetry, f(x) = a(x - 4) + k.
2
Substitute each of the two points into the function to obtain a system of
Substitute (3, -2).
Simplify.
© Houghton Mifflin Harcourt Publishing Company
two linear equations.
2
-2 = a(3 - 4) + k
2
-2 = a(-1) + k
-2 = a + k
-26 = a(6 - 4) + k
Substitute (6, -26).
2
-26 = a(2) + k
Simplify.
-26 = 4a + k
⎧ a + k = -2
The system is ⎨
. Subtract the first equation from the
⎩ 4a + k = -26
second equation, and then solve for a.
3a = -24
a = -8
2
Substitute a = -8 into the first equation and solve for k.
-8 + k = -2
k=6
f(x) = -8(x - 4) + 6
2
Module 3
A2_MTXESE353930_U2M03L2 142
142
Lesson 2
20/02/14 5:21 AM
Writing Quadratic Functions
142
Find the solution to the real-world problem using the given information to write a
quadratic function. If necessary, estimate to answer the question.
INTEGRATE TECHNOLOGY
11. A soccer player who is 27 feet from a goal attempted to kick the
ball into the goal. The ball reached a maximum height of 10 feet
when it was 15 feet from the soccer player. What is the height of
the ball when it reaches the goal? Will the ball go into the goal or
go over it? The goal has a height of 8 feet.
Encourage students to check their work by entering
their functions on a graphing calculator to see
whether their functions produce the parabola with
the given attributes. The table function may be more
useful than the graphing function for a quick check
of values.
The vertex is (15, 10), so h(x) = a(x -15) + 10.
2
Substitute the values of the non-vertex point (0, 0)
into the function and solve for a.
0 = a(0 - 15) + 10
2
0 = a(-15) + 10
-10 = 225a
2
a = -_
45
2
2 (27 - 15) 2 + 10
h(x) = -_
45
2 (144) + 10
= -_
45
= -6.4 + 10
© Houghton Mifflin Harcourt Publishing Company • Image Credits: (t)
©Fotokostic/Shutterstock; (b) ©Jason Hawkes/Getty Images
= 3.6
The ball will be at a height of 3.6 feet when it reaches the goal, so the ball will go into
the goal.
12. A surveyor finds that the cross-section of the bottom of a
circular pond has the shape of a parabola. The pond is 24 feet
in diameter. The middle of the pond is the deepest part at
8 feet deep. At a point 2 feet from the shore the water is 3 feet
deep. How deep is the pond at a point 6 feet from the shore?
The vertex is (12, 8), so d(x) = a(x - 12) + 8.
2
Substitute (2, 3) into the function and solve for a.
3 = a(2 - 12) + 8
2
3 = a(-10) + 8
2
-5 = 100a
1
a = -_
20
1(
d(x) = -___
6 - 12) + 8
2
20
1( )
36 + 8
= -___
20
= -1.8 + 8
= 6.2
The pond is 6.2 feet deep at a point 6 feet from the shore.
Module 3
A2_MTXESE353930_U2M03L2.indd 143
143
Lesson 3.2
143
Lesson 2
2/22/14 5:30 AM
13. A horizontal pedestrian bridge is supported from a parabolic arch. The bridge goes
above a roadway that is 40 feet wide. At ground level, the main span of the bridge
is 90 feet wide. At the edge of the roadway, 25 feet from where the arch touches the
ground, the arch is 16 feet high. How tall is the arch at its tallest point?
PEERTOPEER DISCUSSION
Ask students to discuss with a partner how the
symmetry of a parabola would enable them to find a
non-vertex point that could be used to check their
equations. Since the parabola is symmetric about
the axis of symmetry, you can find the reflection
across the axis of symmetry of one of the given
points, and test to see if its coordinates satisfy the
equation.
The axis of symmetry, x = 45, gives the value of h, so f(x) = a(x - 45) + k.
2
Substitute each of the two points for x and f(x) to obtain a system of two linear equations in
two unknowns.
0 = a(90 - 45) + k
2
0 = a(45) + k
2
Substitute (90, 0).
Simplify.
0 = 2025a + k
16 = a(25 - 45) + k
2
16 = a(-20) + k
2
Substitute (25, 16).
Simplify.
16 = 400a + k
⎧2025a + k = 0
The system is ⎨
.
⎩ 400a + k = 16
Solve the first equation for k to get k = -2025a.
Substitute for k in the second equation
and solve for a.
16 = 400a - 2025a
16 = -1625a
16
_
= a
-
1625
16 into the first equation.
Substitute a = -_
1625
⎛ 16 ⎞
⎟ +k=0
2025⎜-_
Substitute.
⎝ 1625 ⎠
⎞
⎛ _
16
⎟
Solve for k.
k = -2025⎜⎝ 1625 ⎠
1296
k = _
Simplify.
65
1296
The arch is _ feet, or approximately 19.9 feet, tall at its tallest point.
65
© Houghton Mifflin Harcourt Publishing Company
Module 3
A2_MTXESE353930_U2M03L2 144
144
Lesson 2
20/02/14 5:21 AM
Writing Quadratic Functions
144
14. A golfer needs to hit a ball a distance of 500 feet, but there is a 60-foot tall tree
that is 100 feet in front of the point where the shot needs to land. Given that the
maximum height of the shot is 120 feet, and that the intended distance of 500
feet is reached, by how much did the ball clear the tree?
The vertex is (250, 120). Substitute these values to get h(x) = a(x - 250) + 120.
2
Substitute the values of the non-vertex point (0, 0) and solve for a.
0 = a(0 - 250) + 120
2
0 = a(- 250) + 120
2
-120 = 62,500a
6
a = -_
3125
6 (400 - 250) 2 + 120
h(x) = -_
3125
6 (22, 500) + 120
= -_
3125
= -43.2 + 120
= 76.8
The ball will be at a height of 76.8 feet when it is at the distance of the tree so the ball will
clear the tree by 16.8 feet.
Explain whether each function fits the given data.
The parabola opens upward and has x = 8 as an axis of symmetry.
15. ƒ(x) = 2(x - 8) 2 + k
2
16. ƒ(x) = -4(x - 8) + k
Since the value of a is positive and the
correct axis of symmetry is used, this
function fits the data.
1 (x + 8) + k
17. ƒ(x) = _
2
Since an incorrect axis of symmetry is
used, this function does not fit the data.
2
18. ƒ(x) = 71(x - 8) + k
2
19. ƒ(x) = -9(x - 8) + k
2
20. ƒ(x) = (x - 8) + k
© Houghton Mifflin Harcourt Publishing Company
2
A2_MTXESE353930_U2M03L2 145
Lesson 3.2
Since the value of a is positive and the
correct axis of symmetry is used, this
function fits the data.
Since the value of a is negative, this
function does not fit the data.
Module 3
145
Since the value of a is negative, this
function does not fit the data.
Since the value of a is positive and the
correct axis of symmetry is used, this
function fits the data.
145
Lesson 2
1/12/15 4:24 PM
AVOID COMMON ERRORS
H.O.T. Focus on Higher Order Thinking
21. Explain the Error A student was given the following information and asked to find
the function of the parabola in vertex form. Find the student’s error and the correct
function.
Students may make errors when sketching the graph
that models the real-world situation, misidentifying
the measurements that should be used for the
horizontal and vertical axes. Prompt them to analyze
the ordered pairs that are given in the problem, and
use them to identify the type of measurement
indicated by each coordinate. Show them how to use
this information to set up their graphs, and how this
will enable them to interpret the graphs (and, hence,
the functions) correctly.
axis of symmetry: x = 5; points: (3, -20), (6, 1); direction of opening: downward
From the axis of symmetry, ƒ(x) = a(x + 5) + k.
ƒ(x) = a(x + 5) 2 + k
-20 = a(3 + 5) 2 + k
Substitute (3, -20).
-20 = a(8) + k
Simplify.
2
-20 = 64a + k
ƒ(x) = a(x + 5) 2 + k.
1 = a(6 + 5) 2 + k
Substitute (6, 1)
1 = a(11) + k
Simplify.
2
1 = 121a + k
⎧ 64a + k = -20
The system is ⎨
-57a = -21
⎩ 121a + k = 1
.
7
a=_
19
7
Substitute a = __
into the second equation and solve for k.
19
( )
7 +k=1
121 _
19
( )
© Houghton Mifflin Harcourt Publishing Company
7
k = 1 - 121 _
19
828
k = -_
19
7 x+5 2-_
828
ƒ(x) = _
(
)
19
19
The student used x = -5 for the axis of symmetry and wrote
2
f(x) = a(x + 5) + k.
The correct work is shown.
From the axis of symmetry, f(x) = a(x - 5) + k.
2
-20 = a(3 - 5) + k
2
-20 = a(-2) + k
2
Substitute (3, -20).
Simplify.
-20 = 4a + k
Module 3
A2_MTXESE353930_U2M03L2.indd 146
146
Lesson 2
2/22/14 5:40 AM
Writing Quadratic Functions
146
1 = a(6 - 5) + k
2
JOURNAL
1 = a(1) + k
2
Have students describe how knowing various
combinations of attributes of the graph of a quadratic
function can help to determine the rule for the
function.
Substitute (6, 1).
Simplify.
1=a+k
⎧4a + k = -20
The system is ⎨
.
⎩ a+k=1
3a = -21
a=-7
Substitute a = -7 into the first equation and solve for k.
-7 + k = 1
f(x) = -7(x - 5) + 8
2
k=8
22. Make a Conjecture Is it possible to find a unique quadratic function given only
the axis of symmetry, a non-vertex point, and the average rate of change between the
point and vertex? Explain why or why not.
Yes; the average rate of change will give the direction of the opening. Also,
the vertex can be determined using the non-vertex point and the average
rate of change between the point and the vertex.
23. Communicate Mathematical Ideas Is there a unique quadratic function with
points (1, 5) and (5, 5) and axis of symmetry x = 3? Why or why not?
© Houghton Mifflin Harcourt Publishing Company
No; the points are images of each other over the axis of symmetry. This
situation results in a system of two equations in two unknowns whose
solution is the identity 0 = 0. Therefore, there are an infinite number of
quadratic functions that satisfy the given conditions. The vertex can be
any point on the axis of symmetry.
Module 3
A2_MTXESE353930_U2M03L2 147
147
Lesson 3.2
147
Lesson 2
20/02/14 5:21 AM
Lesson Performance Task
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Reasoning
At a test track, the time it takes a driver to stop a car once a signal is given is being measured.
The car travels down a straight section of track at a known speed, and a light flashes. The
driver applies the brakes, and the distance it takes for the car to stop is measured. The stopping
distance can be modeled as a quadratic function of speed. If the quadratic model has an axis of
symmetry of s = -10, what is the function modeling the stopping time if the driver stops in 75
feet if he is traveling 30 miles per hour?
Discuss with students the fact that although the axis
of symmetry is s = -10, values of s that are less than
0 are not values in the domain for the real-world
application because the speed cannot be a negative
number. Help them to see that the axis of symmetry
determines the symmetry of the parabola, and
provides information that helps determine the rule
for the quadratic function.
Now, take the function you found and convert it to standard form, if necessary. The terms all
represent different aspects of the situation, with the constant term representing road conditions,
the linear term representing the driver’s reaction time, and the quadratic term representing the
condition of the car’s brakes and tires. Examine the effects of varying the constant term and the
coefficients of the variable terms. What could these changes represent?
Another point on the curve would be (0, 0) because a car that isn’t moving has no need of
2
stopping. The function d(s) = a(s - h) + k will model d, the stopping distance in terms of
speed, s. The axis of symmetry s = -10 shows that h = -10. Now create and solve the system
using the points (0,0) and (30, 75).
d(s) = a(s + 10) + k
INTEGRATE TECHNOLOGY
2
0 = a(0 + 10) + k
2
0 = a(10) + k
2
d(s) = a(s + 10) + k
Substitute (0, 0).
2
75 = a(30 + 10) + k
2
Simplify.
75 = a(40) + k
0 = 100a + k
2
100a + k = 0
Solve for a.
1600a + k = 75
1600a - 100a = 75
1500a = 75
1
a=_
20
1.
Solve for k at a = _
20
k = -100a
1
k = -100 _
20
k = -5
( )
model:
1 s + 10 = d(s) = _
1 s2 + s
d(s) = _
)
(
20
20
2
Adding a constant term could represent a poor road surface. Changing the linear coefficient
could represent very slow or very fast reaction time. Changing the quadratic coefficient could
represent low-performing brakes or tires.
Module 3
148
© Houghton Mifflin Harcourt Publishing Company
k = -100a
Simplify.
75 = 1600a + k
⎧ 100a + k = 0
This gives the system ⎨
.
⎩ 1600a + k = 75
Solve the first equation for k.
Students may want to use the table feature of a
graphing calculator to compare their function values
for the same value of s. For example, if students add a
constant term or a coefficient to model a change in
road conditions or reaction time, they can enter both
functions into the calculator and use the table to see
how these factors affect stopping distance for a car
traveling at a particular speed.
Substitute (30, 75).
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 2
EXTENSION ACTIVITY
A2_MTXESE353930_U2M03L2.indd 148
Stopping distance is based on physics and is determined by a variety of conditions,
forces, and reaction times. Different road surfaces have different coefficients of
friction between the tires and the road. Students can research the different types
of road surfaces and how they affect stopping distances. One type of surface may
make it easier for a vehicle to stop quickly in dry conditions, but not stop as
quickly in wet or icy conditions, compared to another surface. Heavier vehicles,
such as tractor-trailer rigs (eighteen-wheelers), take longer to stop than lighter
vehicles. Have students present their findings in small groups.
1/10/15 5:11 PM
Writing Quadratic Functions
148