LESSON
3.3
Name
Fitting Quadratic
Functions to Data
Class
3.3
Date
Fitting Quadratic
Functions to Data
Essential Question: How can you find the equation of a quadratic function to model data?
Texas Math Standards
A2.8.B: Use regression methods available through technology to write … a quadratic function
… from a given set of data. Also A2.4.E, A2.8.C
The student is expected to:
Explore
A2.8.B
Investigating Quadratic Function Models for Data
You have worked in various situations with quadratic functions in standard form, ƒ(x) = ax + bx + c, and in
2
vertex form, ƒ(x) = a(x - h) + k where (h, k) gives the coordinates of the vertex of the graph. You can also write a
quadratic function in intercept form, ƒ(x) = a(x - x 1)(x - x 2) where x 1 and x 2 give the x-intercepts of the graph.
2
Use regression methods available through technology to write … a
quadratic function … from a given set of data. Also A2.4.E, A2.8.C
Mathematical Processes
When you have a set of data values that reasonably can be modeled by a quadratic function, you may use any of
the above forms to write the function.
A2.1.F

A science class collects the following lab data. The students plot
f(x)
36
the points as shown. The teacher then asks students to find a
32
reasonable quadratic model for the data.
28
24
3
4
5
6
7
8
9
10
11
12
20
x
16
18
28
31
33
33
27
21
13
2
f(x) 0
12
8
4
The student is expected to analyze mathematical relationships to connect
and communicate mathematical ideas.
Language Objective
2.I.4, 2.I.5, 3.D.1, 3.D.2, 5.B.2, 5.G.3
Work with a partner to compare and contrast quadratic functions in
vertex form and intercept form.
0
Possible answer: by estimating the vertex and
another point to find a model in vertex form,
estimating x–intercepts and another point to find a
model in intercept form, and using technology to
find a regression model in standard form
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
Essential Question: How can you find
the equation of a quadratic function to
model data?
Resource
Locker
Angelíque estimates the vertex to be (7.5, 34), so that
2
ƒ(x) = a(x - 7.5) + 34. She chooses the point (4, 18), substitutes
its coordinates for (x, ƒ(x)) in the equation, and solves for a to
2
obtain the model ƒ(x) = -1.31(x - 7.5) + 34. Her graph is shown.
36
32
28
24
20
16
12
8
4
The vertex of the model is at (7.5, 34), so the axis of
symmetry is x = 7.5 and the maximum is
34
.
The x-intercepts of the graph of the model are
about 2.4 and 12.6 .
0
x
1 2 3 4 5 6 7 8 9 10 11 12
f(x)
x
1 2 3 4 5 6 7 8 9 10 11 12
Is the model a good fit for the data? Explain.
Possible answer: The model is a fairly good fit. It is very close to the data values near the
vertex and passes through the chosen point. The graph of the model looks to be a little
too wide, however, as it passes outside the majority of data points.
PREVIEW: LESSON
PERFORMANCE TASK
Module 3
ges
EDIT--Chan
DO NOT Key=TX-A
Correction
must be
Lesson 3
149
gh “File info”
made throu
Date
Class
adratic
Fitting Qu to Data
Functions
Name
3.3
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Resour
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Locker
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A2.4.E, A2.8.C
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A2.8.B: Use
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given set
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2
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Explore
quadratic
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gives the
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32
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28
class collec
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A science
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24
as shown
the data.
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20
12
quadratic
11
reasonable
10
16
9
8
2
7
13
12
6
21
5
x
27
4
33
8
3
33
x
31
28
4
12
18
8 9 10 11
f(x) 0
5 6 7
0 1 2 3 4
Quest
Essential
A2_MTXESE353930_U2M03L3.indd 149
HARDCOVER PAGES 107122
Turn to these pages to
find this lesson in the
hardcover student
edition.

34), so that substitutes
to be (7.5,
18),
the vertex
point (4,
estimates
chooses the and solves for a to
2
.
Angelíque
+ 34. She
on,
(x - 7.5)
graph is shown
the equati
ƒ(x) = a
(x, ƒ(x)) in (x - 7.5)2 + 34. Her
nates for
its coordi
x) = -1.31
model ƒ(
obtain the
f(x)
36
32
28
24
20
16
12
8
4
x
axis of
34), so the
l is at (7.5,
12
34 .
of the mode
8 9 10 11
5 6 7
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The x-inte
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.
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2.4
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Explain.
close to
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It is very
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Is the mode
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The graph
answer:
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Lesson 3
too wide,
Publishin
y
g Compan
View the online Engage. Discuss the seasonal aspect
of the hole in the ozone layer above the Antarctic and
why a quadratic function might model its size (area)
over time. Then preview the Lesson
Performance Task.
© Houghto
n Mifflin
Harcour t
The vertex
149
Module 3
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A2_MTXE
149
Lesson 3.3
L3 149
0_U2M03
20/02/14
3:40 AM
1/10/15 6:41 PM
B
Beto observes the intercept at x = 3, and uses x = 12 as an
approximation for the other x-intercept, so that
ƒ(x) = a(x - 3)(x - 12). He chooses the point (7, 33),
substitutes its coordinates for (x, ƒ(x)) in the equation, and
solves for a to obtain the model ƒ(x) = -1.65(x - 3)(x - 12).
His graph is shown.
36
32
28
24
20
16
12
8
4
f(x)
EXPLORE
Investigating Quadratic Function
Models for Data
x
0
1 2 3 4 5 6 7 8 9 10 11 12
INTEGRATE TECHNOLOGY
The intercepts of the model are 3 and 12.
Discuss the fact that, just as for linear
functions, graphing calculators also calculate
a correlation value, in this case R 2, to determine how
well a quadratic function fits the data. To see this
calculation, you may need to enable the correlation
value feature on the device.
x +x
1
2
The axis of symmetry of a parabola is halfway between the x-intercepts, or at x = ______
,
2
3 + 12
so the model’s axis of symmetry is x = ___________
= 7.5 .
2
Because the vertex is on the axis of symmetry, the model’s maximum is
( )
(
)(
)
ƒ 7.5 = -1.65 7.5 - 3 7.5 - 12 . So, to the nearest tenth, the maximum is 33.4 .
Is the model a good fit for the data? Explain.
Possible answer: The model is a good fit. It is extremely close to the data values near the
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Reasoning
vertex, passes through or very near the intercepts, and passes inside and outside about
the same number of data points.
C
Charla enters the ordered pairs into her graphing calculator
and performs quadratic regression to find the equation of the
best-fitting quadratic model. The result is approximately
ƒ(x) = -1.56x 2 + 23.1x - 51.6. The graph is shown.
0
Here are the graphs of the models all together:
36
32
28
24
20
16
12
8
4
0
Module 3
150
Ask students to consider situations in which an
approximation of a quadratic model, as opposed to
an exact model, would be sufficient.
f(x)
x
1 2 3 4 5 6 7 8 9 10 11 12
f(x)
B
A
C
B, C
© Houghton Mifflin Harcourt Publishing Company
Using the statistics calculation features on her calculator,
Charla finds that the x-intercepts are about 2.7 and 12.1,
the axis of symmetry is about x = 7.4, and the maximum
is about 33.9.
36
32
28
24
20
16
12
8
4
A
x
1 2 3 4 5 6 7 8 9 10 11 12
Lesson 3
PROFESSIONAL DEVELOPMENT
A2_MTXESE353930_U2M03L3 150
Learning Progressions
20/02/14 3:40 AM
Students have learned how to fit a linear function to data using both
approximation techniques and linear regression. In this lesson, students expand
their knowledge of these techniques to nonlinear data. Although it is given that
the data sets in the examples in this lesson are quadratic, ultimately students will
be expected to be able to analyze a set of data and determine what type of function
(linear, quadratic, exponential, etc.) can best be used to model the data.
Fitting Quadratic Functions to Data
150
Give a quick comparison of the graphs of the models. How are they alike? How are they different?
QUESTIONING STRATEGIES
Possible answer: The axes of symmetry of all three models are very close to one another.
What are the advantages of having a quadratic
function written in vertex form or in intercept
form? In vertex form, you can read the coordinates
of the vertex directly from the equation. In intercept
form, you can read the x–intercepts of the function
directly from the equation.
The maximums are also close to the same number, with the maximum for Angelíque’s
model a little more than the other two. The graphs for Beto and Charla’s graphs are almost
the same, especially on their right sides, with the left side of Charla’s model extending a
little wider to the left. The graph of Angelíque’s model lies a little outside the other two, so
its x-intercepts are a little farther from the axis of symmetry.
Reflect
1.
Look at the graph of Angelíque’s model. Keeping her original estimate for the vertex, what might she have
done differently that would have made her model closer to the other two?
Possible answer: If she had used a different second point, say, (3, 0) or (11, 13), her model
would have been closer, since the other models pass through or very near those points.
2.
Discussion What are some reasons that you might use each of the methods in this Explore to find a
quadratic model for a data set?
Possible answer: You might use the vertex or intercept form models if you want a quick,
informal model of a data set, if you want to make general observations, or if you believe
© Houghton Mifflin Harcourt Publishing Company
your estimates for the intercepts or the vertex are likely to be fairly accurate. If you need
to make predictions that are as accurate as possible and are sure that a quadratic model
is appropriate, using technology to perform quadratic regression would be appropriate.
This is especially true when you do not have data points in the area of the vertex or the
x-intercepts.
Module 3
151
Lesson 3
COLLABORATIVE LEARNING
A2_MTXESE353930_U2M03L3 151
Peer-to-Peer Activity
Have students work in pairs. One student in each pair writes a quadratic function
and gives the partner 4 points that lie on the function’s graph, two on either side of
the axis of symmetry. The partner then plots the points, and approximates a
quadratic model for the data (without using technology). Ask students to compare
the model with the original function, and discuss any discrepancies.
151
Lesson 3.3
20/02/14 3:40 AM
Explain 1
Roughly Fitting a Quadratic Function In Vertex
Form To Data
EXPLAIN 1
When you can make a reasonable estimate of the vertex for a quadratic model on a data plot, you can find an
equation for the model. First, visually estimate the axis of symmetry. It may help to sketch parabolic curves
among the data points. Use your estimate to approximate the maximum or minimum of the model curve.
Roughly Fitting a Quadratic Function
in Vertex Form to Data
Using your estimate of the vertex (which may be a data point) and the coordinates of a second point, solve for
the parameter a to find a model in vertex form. The second point does not have to be a data point, but should
be a point that you expect to lie on or very near the graph of an appropriate model.
Example 1

COLLABORATIVE LEARNING
Find an approximate quadratic function model for the graphed data by
estimating the coordinates of the vertex and using one other point.
Ask students to share how they go about estimating
the locations of the axis of symmetry and the vertex.
Have several students demonstrate their methods
using different sets of data.
The yearly average high temperature in Oakdale is 60°F. The ordered pairs (m, D) give the
month of the year m (where January is month 1) and the number of degrees D (Fahrenheit)
by which the average high temperature for the months May through October exceeds the
yearly average high.
So for example, in June (m = 6), the average high is 19°
above the yearly average, or 60 + 19 = 79°. The graph
of the data is shown.
Estimate the vertex.
Notice that the points for August and September are a little
higher than the points for May and June. This indicates that
the best estimate for the axis of symmetry is a little to the
right of m = 7, and so the model’s maximum will be a little
more than 25. Estimate (7.2, 25.5) for the vertex. Use this
with another point, say (10, 1), to find a model.
27 D
24
21
18
15
12
9
6
3
0
m
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
ƒ(x) = a(x - h) + k
2
ƒ(x) = a(x - 7.2) + 25.5
2
1 = a(10 - 7.2) + 25.5
2
-24.5 = a(2.8)
2
© Houghton Mifflin Harcourt Publishing Company
Write the vertex form.
Substitute (7.2, 25.5) for (h, k).
Substitute (10, 1) for (x, ƒ(x)).
Solve for a.
Departure from Avg. (°F)
(5, 10), (6, 19), (7, 25), (8, 21), (9, 14), (10, 1)
-24.5 = a
_
(2.8) 2
a = -3.125
So, ƒ(x) = -3.125(x - 7.2) + 25.5.
2
An approximate model is D = -3.1(m - 7.2) + 25.5.
2
Departure from Avg. (°F)
The graph is shown.
27 D
24
21
18
15
12
9
6
3
0
Module 3
152
m
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
Lesson 3
DIFFERENTIATE INSTRUCTION
A2_MTXESE353930_U2M03L3.indd 152
Multiple Representations
1/10/15 6:41 PM
Students may not readily make the connection between a data set and the function
that models the situation. Help them to understand that each given data point
represents the pairing of an output value (y–coordinate) to a specific input value
(x–coordinate) of a particular relation, and that the process of fitting a function to
the data is the search to identify the function that best approximates the rule that
produces those ordered pairs.
Fitting Quadratic Functions to Data
152
B
Engineers conduct a traffic study
120 V
on a road that has more traffic
105
than it was designed for. At
90
75
10-minute intervals from 7:30 to
60
9:10 AM, they record how many
45
vehicles pass a given point in 1
30
minute. The ordered pairs (t, V)
15
t
give the results, where t indicates
0
1
2
3
4
5
6
7
8
9
10
11
the time of recording (0 for 7:30,
-15
1 for 7:40, and so on) and V
10-min intervals after 7:30 AM
indicates how many vehicles in
excess of the designed maximum
passed in that minute (a negative sign means the number was below the maximum).
Excess vehicles
QUESTIONING STRATEGIES
What information does the graph of the data
points tell you about the graph of the
function? It tells you the direction of opening, and
gives you a general idea about where the vertex will
be located.
(0, –28) (1, 40), (2, 78), (3, 90), (4, 111), (5, 99), (6, 106), (7, 78), (8, 70), (9, 54), (10, 14)
From looking at the graph, reasonable estimates for the
vertex may vary quite a bit. From the overall symmetry and
considering the fact that the engineers would not want to
underestimate the maximum, a reasonable estimate is
[(4, 110)/(5, 115)/(6, 105)].
Pick a second point not too near the vertex that seems to fit
the overall trend, say (1, 40), to find a model.
Excess vehicles
Estimate the vertex.
120 V
105
90
75
60
45
30
15
t
0
1
2
3
4
5
6
7
8
9
10
11
-15
© Houghton Mifflin Harcourt Publishing Company · Image Credits: ©SP-Photo/
Shutterstock
10-min intervals after 7:30 AM
ƒ(x) = a(x - h) + k
2
Write the vertex form.
(
Substitute (
Substitute
)
40 ) for (x, ƒ x ).
(
40 = a
( )
1 ,
Solve for a.
(
ƒ(x) = a x -
5 , 115 for (h, k).
(
1
-75 = a -4
-75
______ = a
5
) + 115
2
)
2
- 5 + 115
)
2
16
(
So, ƒ(x) = -4.6875 x -
5
a = -4.6875
) + 115 .
2
An approximate model (to the nearest tenth) is V = -4.7( t - 5 )2 + 115 .
The graph is shown.
Module 3
153
Lesson 3
LANGUAGE SUPPORT
A2_MTXESE353930_U2M03L3.indd 153
Connect Vocabulary
Have each pair of students analyze a quadratic function, one in vertex form and
one in intercept form. Have students write on note cards the information they can
get from each form: the vertex from vertex form; minimum and maximum values
from vertex form; x–intercepts from intercept form; and so on.
153
Lesson 3.3
1/10/15 6:41 PM
Reflect
3.
In Part A, the model looks like a good fit for the warmer half of the year. How might a quadratic model for
the cooler half of the year differ?
It would probably have about the same shape, but would open upward. It would basically
be a reflection across the x-axis that is horizontally shifted 6 months.
4.
In Part B, the model looks like a reasonably good fit for the morning rush hour period. Is there another
portion of the day that might be modeled by a similar curve? Explain. Why would the engineers be most
interested in the models for these portions of the day?
Yes; the traffic will likely increase again in a similar pattern for the afternoon rush hour
period, as people go back home. The engineers would be most interested in times when
the traffic exceeds the intended safe capacity, since the other time periods are not a
problem.
Your Turn
5.
The yearly average low temperature in Houston, Texas is 60°F.
The ordered pair (m, D) gives the month m (counting October as m = 1, November as m = 2, and so on),
and the number of degrees D (Fahrenheit) by which the average low temperature
for the months October through April departs from the yearly average low.
Possible answer, using (4, –16) as the vertex and
(1, 1):
f(x) = a(x - 4) - 16
2
1 = a(1 - 4) - 16
2
17 = a(-3)
2
m
D
1 2 3 4 5 6 7 8 9
Month (Oct. = 1)
17
___
=a
9
a ≈ 1.889
0
-2
-4
-6
-8
-10
-12
-14
-16
f(x) = 1.889(x - 4) - 16.
© Houghton Mifflin Harcourt Publishing Company
Find an approximate quadratic function model for the graphed
data by estimating the coordinates of the vertex and using one
other point. Then sketch your graph for D ≤ 0 on the grid.
Departure from Avg. (°F)
(1, 1), (2, –8), (3, –15), (4, –16), (5, –14), (6, –6), (7, 0)
So for example, in January (m = 4), the average low is 16° below the yearly average,
or 60 - 16 = 44°. The graph of the data is shown.
2
An approximate model is D = 1.9(m - 4) -16.
Module 3
A2_MTXESE353930_U2M03L3.indd 154
2
154
Lesson 3
1/10/15 6:41 PM
Fitting Quadratic Functions to Data
154
Explain 2
EXPLAIN 2
Roughly Fitting a Quadratic Function in Intercept
Form to Data
Finding a model in intercept form is very similar to finding a model in vertex form, except that this time you begin
with estimates for the x-intercepts of the model.
Roughly Fitting a Quadratic Function
in Intercept Form to Data
Example 2

AVOID COMMON ERRORS
Find an approximate quadratic function model for the graphed data by
estimating the x-intercepts and using one other point.
Use the scenario from Example 1, Part A for your model. The data and graph are repeated here.
(5, 10), (6, 19), (7, 25), (8, 21), (9, 14), (10, 1)
Students may be confused as to the zeros of a
function whose vertex lies on the x–axis. Explain that
the x–coordinate of the vertex is actually a repeated
zero of the function, and that its value should be used
for both x 1 and x 2 in the intercept form of
the function.
On the right side of the graph, you can see that the
x-intercept is a tiny bit to the right of 10. Estimate 10.1.
On the left side, it is apparent that the graph is negative
when x is 4, so estimate the intercept as a little farther to
the right, say 4.3.
Notice that the axis of symmetry of the model is
4.3 + 10.1
x = ________
= 7.2, which is the same as in the vertex
2
model in Example 1.
Departure from Avg. (°F)
Estimate the x-intercepts.
27 D
24
21
18
15
12
9
6
3
0
Use the intercept estimates, 4.3 and 10.1, along with
another point, say (7, 25), to write a model.
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
ƒ(x) = a(x - x 1)(x - x 2)
ƒ(x) = a(x - 4.3)(x - 10.1)
25 = a(7 - 4.3)(7 - 10.1)
Write the intercept form.
Substitute 4.3 for x 1 and 10.1 for x 2.
Substitute (7, 25) for (x, ƒ(x)).
Solve for a.
m
25 = a(2.7)(-3.1)
25
_________
=a
(2.7)(-3.1)
a ≈ -2.987
© Houghton Mifflin Harcourt Publishing Company
So, ƒ(x) = -2.987(x - 4.3)(x - 10.1).
An approximate model is D = -3.0(m - 4.3)(m - 10.1).
Departure from Avg. (°F)
The graph is shown.
27 D
24
21
18
15
12
9
6
3
0
m
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
Module 3
A2_MTXESE353930_U2M03L3 155
155
Lesson 3.3
155
Lesson 3
20/02/14 3:40 AM
Use the scenario from Example 1, Part B for your model.
The data and graph are repeated here.
(0, –28) (1, 40), (2, 78), (3, 90), (4, 111), (5, 99), (6, 106),
(7, 78), (8, 70), (9, 54), (10, 14)
Excess vehicles
B
120 V
105
90
75
60
45
30
15
t
0
1 2 3 4 5 6 7 8 9 10 11
-15
QUESTIONING STRATEGIES
How can you convert a quadratic function
that is written in intercept form to vertex
form? You can multiply the factors in intercept
form, and then use completing the square to write
the resulting polynomial in vertex form.
10-min intervals after 7:30 AM
Excess vehicles
Estimate the x-intercepts.
Why would this be useful? Because in
addition to knowing the intercepts of the
graph, you would also know the vertex and could
easily determine the range of the function.
120 V
105
90
75
60
45
30
15
t
0
1 2 3 4 5 6 7 8 9 10 11
-15
INTEGRATE TECHNOLOGY
Students can use a graphing calculator to
check their work. They can enter and graph
their functions, and check that the resulting graph
contains the given x–intercepts and the additional
point they used to find a. Note that the graph may or
may not pass through any additional data points.
10-min intervals after 7:30 AM
The points do not all clearly lie along a single curve, but reasonable estimates are
[0 and 11/0.5 and 10/1 and 10].
Pick a second point that is not near the intercepts, say (6, 106), to write a model.
ƒ(x) = a(x - x 1)(x - x 2)
Substitute 0.5 for x 1 and 10 for x 2.
ƒ(x) = a x - 0.5
Substitute
(
)
6 , 106 for (x, ƒ(x)).
Solve for a.
(
So, ƒ(x) = -4.818 x - 0.5
)(x -
10
106
(
)(x - 10 )
= a( 6 - 0.5)( 6 - 10)
= a( 5.5 )( -4 )
106
106 = a
_
-22
a ≈ -4.818
).
An approximate model (to the nearest tenth) is V = -4.8(t - 0.5)(t - 10) .
The graph is shown.
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156
© Houghton Mifflin Harcourt Publishing Company
Write the intercept form.
Lesson 3
1/10/15 6:41 PM
Fitting Quadratic Functions to Data
156
Your Turn
6.
Fitting a Quadratic Function to Data
Using Technology
Use the scenario from Your Turn 5. Find an approximate quadratic function model for the graphed data by
estimating the x-intercepts and using one other point. Then sketch your graph for D ≤ 0 on the grid.
Possible answer, using 1.1 and 7 as the x-intercepts and
(4, -16) as the vertex:
f(x) = a (x - x 1)(x - x 2)
f(x) = a(x - 1.1)(x - 7)
-16 = a(4 - 1.1)(4 - 7)
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Modeling
-16 = a(2.9)(-3)
-16
____
=a
-8.7
a ≈ 1.839
0
-2
-4
-6
-8
-10
-12
-14
-16
m
D
1 2 3 4 5 6 7 8 9
Month (Oct. = 1)
So, f(x) = 1.839(x - 1.1)(x -7).
Explain to students that in mathematics a regression is
a function that closely models a set of data. A
calculator produces a regression by minimizing the
vertical distance between each data point and the
function’s graph.
Departure from Avg. (°F)
EXPLAIN 3
An approximate model is D = 2.0(m - 1.1)(m - 7).
Explain 3
Fitting a Quadratic Function To Data Using
Technology
You have used a graphing calculator to find the equation of a line of best fit by entering data and then performing
linear regression. You can also use a graphing calculator or software application to find a quadratic curve of best fit
for a set of data by performing quadratic regression.
Use the quadratic regression feature on a graphing calculator to find an
approximate quadratic function model for the graphed data.
Example 3
A2_MTXESE353930_U2M03L3.indd 157
Lesson 3.3
y
5
10
6
19
7
25
8
21
9
14
10
1
27 D
24
21
18
15
12
9
6
3
0
m
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
Module 3
157
x
Departure from Avg. (°F)
Use the scenario from Example 1, Part A for your model. The data are repeated in the table.
© Houghton Mifflin Harcourt Publishing Company

157
Lesson 3
3/19/15 2:16 PM
Using the statistics editor on your calculator, enter the data, using
List 1 for the month and List 2 for the number of degrees.
QUESTIONING STRATEGIES
How does the regression model produced by
the calculator compare to the models obtained
in the previous examples? The equation produced
by the calculator is written in standard form. In the
previous examples, the equation was written in
vertex form or in intercept form.
From the statistics calculations menu, select quadratic regression.
Make sure the calculator is using List 1 for the x-values and List 2
for the y-values. The result is shown.
How can you find the x–intercepts of the
graph of the regression equation? You can
graph the equation and use the zero feature in the
calculate menu.
Using the statistics calculation features on a graphing
calculator, you can find that the x-intercepts are about 4.3
and 10.0, the axis of symmetry is about x = 7.2, and the
maximum is about 23.7.
The graph is shown.
Departure from Avg. (°F)
An approximate model is D = -2.9m 2 + 41.6m - 125.6.
27
24
21
18
15
12
9
6
3
0
D
How can you find the vertex of the graph of
the regression equation? You can graph the
equation and use the maximum or minimum
feature in the calculate menu.
m
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
Use the scenario from Example 1, Part B for your model.
The data are repeated in the table.
y
0
-28
1
40
2
78
3
90
4
111
5
99
6
106
7
78
8
70
9
54
10
14
Module 3
A2_MTXESE353930_U2M03L3.indd 158
120 V
105
90
75
60
45
30
15
t
0
1
2
3
4
5
6
7
8
9
10
11
-15
10-min intervals after 7:30 AM
158
© Houghton Mifflin Harcourt Publishing Company
x
Excess vehicles

Lesson 3
19/03/15 10:23 AM
Fitting Quadratic Functions to Data
158
Using the statistics editor on your calculator, enter the data, using
List 1 for the number of the time interval and List 2 for the excess
number of vehicles.
From the statistics calculations menu, select quadratic regression.
Make sure the calculator is using List 1 for the x-values and List 2 for
the y-values. The result is shown.
An approximate model (to the nearest tenth) is
V = -4.4t + 45.6t - 10.4 .
Graph the quadratic regression model and use the statistics
calculation features on your calculator to find the following
(round to the nearest tenth):
The x-intercepts are about 0.2 and 10.1 .
The axis of symmetry is about x = 5.2 .
The maximum is about
Excess vehicles
2
107.7 .
120
105
90
75
60
45
30
15
0
-15
V
t
1 2 3 4 5 6 7 8 9 10 11
10-min intervals after 7:30 AM
Reflect
7.
A graphing calculator will give you the most accurate quadratic model of a set of data. Can you think of
any disadvantages of using quadratic regression?
Possible answer: The calculator treats all data points equally, so if you are more sure
of some data points or consider some points more important, the calculator will not
adjust the model. Also, just because a calculator will find the best quadratic model
does not guarantee that a quadratic model is the best model, so it may give you a
Your Turn
Find an approximate quadratic function model for the graphed data by estimating the
x-intercepts and using one other point.
8.
Use the scenario from Your Turn 5. Use the quadratic regression
feature on a graphing calculator to find an approximate
quadratic function model for the graphed data. Then sketch
your graph for D ≤ 0 on the grid.
Regression equation: y ≈ 1.857x 2 - 14.786x + 13.714;
An approximate model is D = 1.9m 2 - 14.8m + 13.7.
Departure from Avg. (°F)
© Houghton Mifflin Harcourt Publishing Company
false sense of accuracy.
0
-2
-4
-6
-8
-10
-12
-14
-16
m
D
1 2 3 4 5 6 7 8 9
Month (Oct. = 1)
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A2_MTXESE353930_U2M03L3 159
159
Lesson 3.3
159
Lesson 3
1/28/15 8:58 AM
Explain 4
Solving a Real-World Problem
EXPLAIN 4
Once you have a suitable model for a data set, you can use it to make predictions and draw conclusions for
appropriate intervals of the domain and range. Remember to be careful, however, about applying the model outside
the region of your data, where it may not be appropriate.
Solving a Real-World Problem
One way to make a prediction is simply to substitute a value for a variable in the model and solve for the other
variable, but you may also use the graph of the model.
Example 4

Use the indicated quadratic models to answer the questions.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Communication
The graph shows the models from part A of Examples 1 and 3:
Vertex model: D = -3.1(m - 7.2) + 25.5
Regression model: D = -2.9m 2 + 41.6m - 125.6
Padma is moving to Oakdale from Centerville. In
Centerville, the greatest monthly average high temperature
is 84°, and there is a 105-day period where the average high
is at least 70°.
a. Graph the line D = 24 on the grid. What does this
line represent? (Remember that the average yearly
high temperature in Oakdale is 60°.) What different
conclusions might Padma reach using the graphs of the
two models to compare Oakdale to Centerville? Explain.
Departure from Avg. (°F)
2
27
24
21
18
15
12
9
6
3
D
Ask students how they can explain the fact that the
function models the situation, even though not all of
the data points belong to the function. Ask them to
describe the implication of a data point that lies
farther from the graph of the regression function
than one that lies closer.
m
0
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
CONNECT VOCABULARY
Draw the lines D = 24 and D = 10 on the graph.
27
24
21
18
15
12
9
6
3
0
Have students label the parts of a written quadratic
function (or functions) in standard form, intercept
form, and vertex form.
D
m
1 2 3 4 5 6 7 8 9 10 11
Month (Jan. = 1)
b. The line D = 10 represents a temperature of 60 + 10 = 70°. The models are nearly
identical near 70°. They both predict a period of about four and a half months, or
about 135 days, in Oakdale with an average high of at least 70°. This is about 30 days
longer than the period for Centerville.
Module 3
A2_MTXESE353930_U2M03L3 160
160
© Houghton Mifflin Harcourt Publishing Company
a. The line D = 24 represents a temperature of
60 + 24 = 84°. From the regression model, Padma
could conclude that the hottest part of summer is nearly
identical, since the maximum of the model for Oakdale is
almost exactly 84, which is the maximum for Centerville.
From the vertex model, she could conclude that there
is about a 45-day period where the average high is at
or above 84 degrees in Oakdale, which means Oakdale
is significantly hotter. This is because the graph of the
Oakdale model is above 84° for about a month and a half,
or about 45 days.
Departure from Avg. (°F)
b. Graph the line D = 10 on the grid. What does this line represent? Compare the
predictions of the models for how long the period is in Oakdale with an average high
of at least 70°. How does this compare with Centerville?
Lesson 3
20/02/14 3:39 AM
Fitting Quadratic Functions to Data
160
The graph shows the models from part B of
Examples 1 and 2:
Excess vehicles
B
QUESTIONING STRATEGIES
Vertex model: V = -4.7(t - 5) + 115
2
Which of the equations—vertex form,
intercept form, or the regression equation—is
the most useful for making predictions related to the
situation? The regression equation is likely to be the
most accurate of the three equations, and thus
would produce the most accurate prediction.
Intercept model: V = -4.8(t - 0.5)(t - 10)
The engineers propose a short-term plan to raise the safe
capacity of the road by 60 vehicles per minute. Graph the
line V = 60 on the grid. What does it represent in relation
to the graphs? Do the models differ significantly in the
conclusions they might lead you to make in evaluating the
plan? Explain.
10-min intervals after 7:30 AM
Excess vehicles
First draw the line V = 60 on the graph.
To answer the questions, consider how the line is related
to the number of passing vehicles that are counted as
excess vehicles.
120 V
105
90
75
60
45
30
15
t
0
1 2 3 4 5 6 7 8 9 10 11
-15
120 V
105
90
75
60
45
30
15
t
0
1
2
3
4
5
6
7
8
9
10
11
-15
10-min intervals after 7:30 AM
Possible answer: The line represents a new, higher baseline above which passing
vehicles are counted as excess vehicles in terms of safety. The models would not likely
lead to significantly different conclusions. The vertex model has traffic exceeding
the new baseline at about 7:45 instead of 7:50. Both then have traffic above the new
baseline until about 8:55. The vertex model shows a maximum of about 55 vehicles
above the baseline, while the intercept model has a maximum of about 50 vehicles
effectiveness of the plan.
Your Turn
9.
The line D = -15 along with the vertex-form model from
Your Turn 6 and the quadratic regression model from Your
Turn 8 are shown on the graph. Use the graph to answer the
following questions.
a. What does the line represent?
It represents a temperature that is 15 degrees below
the average, or 60 - 15 = 45°F.
b. The line intersects each model twice. What do these
intersection points represent?
The left point indicates when the average low drops
below 45°F, and the right point indicates when the
average temperature rises back above 45°F.
Module 3
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161
Lesson 3.3
161
Departure from Avg. (°F)
© Houghton Mifflin Harcourt Publishing Company
above the baseline. But these are minor differences as far as evaluating the overall
0
-2
-4
-6
-8
-10
-12
-14
-16
m
D
1 2 3 4 5 6 7 8 9
Month (Oct. = 1)
Lesson 3
1/12/15 4:25 PM
ELABORATE
c. Use the graphs to estimate roughly how long each model predicts the average low
temperatures will be below 45 degrees. Explain how you got your answers.
The regression model predicts a period of about a month, and the intercept model
predicts a period of about two months. Because each grid square has a width of 1 month,
I looked to see the width of the interval for which the model temperature dropped below
the line for 45°F.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Critical Thinking
Elaborate
Ask students to consider when it would be a good
idea to omit one or more data points from a data set
when using quadratic regression to determine a
model for a real-world situation. Lead them to see
that outliers, which may be identified visually, may
distort the data, and hence produce a regression
equation that is not as accurate a model of the
situation as the regression model for the data set
without the outlier(s).
10. How do the data sets modeled in the lesson emphasize the importance of being careful about using the
model to make predictions outside the domain represented by the data?
For the data given, though the models fit the data well, it is clear that for different
domains the shape of an appropriate model will change, that is, the trends shown by the
model will not continue. For example, the departure of the average monthly temperature
compared to the yearly average will reverse direction after crossing the yearly average.
Also, for the traffic model, you would expect a similar curve for the afternoon rush hour,
but between the rush hours, and especially overnight, the traffic trends would change.
11. Discussion Three students have modeled a data set, each using a different one of the methods used in the
Examples. Give two ways they can compare their models. What are the advantages of each?
One way to compare models is to translate between the various forms. For example, you
SUMMARIZE THE LESSON
Given a set of data that is quadratic in nature, what
are some ways to determine a model for the
data? You could plot the data points and
approximate the turning point of the graph of the
function, then use that information to determine
the function in vertex form. You could also
approximate the x–intercepts from a graph of the
data points and use that information to determine
the function in intercept form. You could also use a
graphing calculator to determine a regression
equation.
can multiply to rewrite a vertex form model or intercept form model in standard form, or
complete the square to write a standard form model in vertex form. Translating lets you
compare model parameters directly. Another method is graphing the models together.
This has the advantages of giving a clear, quick overview of the model differences, as
intercepts.
12. Essential Question Check-In When you have a scatter plot for which a quadratic model is appropriate,
how can you decide how to proceed to find a model?
If you are able to make a good estimate for the vertex or x-intercepts, and can observe
the overall pattern well, you can find an approximate model in vertex or intercept form.
When you cannot observe clearly what it happening near the vertex or x-intercepts, when
© Houghton Mifflin Harcourt Publishing Company
well as highlighting important differences such as maximum or minimum points and
you have data for only a restricted part of the curve, or when you need a mathematical
best-fitting model, you can use technology to find a regression equation.
Module 3
A2_MTXESE353930_U2M03L3 162
162
Lesson 3
20/02/14 3:39 AM
Fitting Quadratic Functions to Data
162
Evaluate: Homework and Practice
EVALUATE
• Online Homework
• Hints and Help
• Extra Practice
Find an approximate quadratic model for the data by estimating the coordinates
of the vertex and one other point. Then sketch a graph of the model.
1.
ASSIGNMENT GUIDE
Concepts and Skills
Practice
Explore
Investigating Quadratic Function
Models for Data
f(x)
9
8
7
6
5
4
3
2
1
x
0
2.
Example 1
Roughly Fitting a Quadratic
Function in Vertex Form to Data
Exercises 1–2
Example 2
Roughly Fitting a Quadratic
Function in Intercept Form to Data
Exercises 3–4
Example 3
Fitting a Quadratic Function to Data
Using Technology
Exercises 5–6
Example 4
Solving a Real-World Problem
Exercises 7–9
9
8
7
6
5
4
3
2
1
Use (4, 8.5) as the vertex and (8, 2):
2
f(x) = a(x - 4) + 8.5
2
2 = a(8 - 4) + 8.5
2
-6.5 = a(4)
-6.5 = a
_
16
a ≈ -0.406
2
An approximate model is f(x) = -0.41(x - 4) + 8.5.
1 2 3 4 5 6 7 8 9
Use (4, 0.75) as the vertex and (8, 4):
f(x)
f(x) = a(x - 4) 2 + 0.75
2
4 = a(8 - 4) + 0.75
2
)
(
3.25 = a 4
x
0
1 2 3 4 5 6 7 8 9
3.25
_
= a
16
a ≈ 0.203
An approximate model is f(x) = 0.20(x - 4) + 0.75.
2
Find an approximate quadratic model for the data by estimating the
x-intercepts and one other point. Then sketch a graph of the model.
Ask students to discuss how they can determine
whether a set of data is quadratic in nature or linear,
with outliers distorting its linearity. Data that is
quadratic should approximate a curve that either
increases and then decreases, or decreases and then
increases. It should be possible to approximate an
axis about which the data are symmetric. Data that
are linear approximate a line; the data are generally
increasing or decreasing, but not both.
© Houghton Mifflin Harcourt Publishing Company
PEERTOPEER DISCUSSION
3.
9
8
7
6
5
4
3
2
1
f(x) = a(x - 1.7)(x - 9.2)
8 = a(5 - 1.7)(5 - 9.2)
8 = a(3.3)(-4.2)
x
0
4.
3
2
1
0
-1
-2
-3
-4
-5
Use 1.7 and 9.2 as the x-intercepts and (5, 8):
f(x)
1 2 3 4 5 6 7 8 9
f(x)
x
1 2 3 4 5 6 7 8 9 10
8
__
= a
(3.3)(-4.2)
a ≈ -0.577
An approximate model is f(x) = -0.58(x - 1.7)(x - 9.2).
Use 1.2 and 8.5 as the x-intercepts and (4, -5):
f(x) = a(x - 1.2)(x - 8.5)
-5 = a(4 - 1.2)(4 - 8.5)
-5 = a(2.8)(-4.5)
-5
_
= a
(2.8)(-4.5)
a ≈ 0.397
An approximate model is f(x) = 0.40(x - 1.2)(x - 8.5).
Module 3
Lesson 3
163
A2_MTXESE353930_U2M03L3.indd 163
Exercise
163
Lesson 3.3
2/22/14 6:21 AM
Depth of Knowledge (D.O.K.)
Mathematical Processes
1–4
1 Recall of Information
1.E Create and use representations
5–6
1 Recall of Information
1.C Select tools
7–8
2 Skills/Concepts
1.A Everyday life
9
2 Skills/Concepts
1.F Analyze relationships
10
3 Strategic Thinking
1.G Explain and justify arguments
5.
Use the quadratic regression feature on a graphing calculator to find
an approximate quadratic function model for the data in Evaluate
Exercise 2, found in the table. Then sketch the graph with the graph
of your model, and compare the graphs.
x
2
3
5
7
8
9
y
2
1
1
2
4
7
An approximate model is y = 0.203x 2 - 1.51x + 3.49.
Use the quadratic regression feature on a graphing calculator
to find an approximate quadratic function model for the data
in Evaluate Exercise 4, found in the table. Then sketch the
graph with the graph of your model, and compare the graphs.
x
1
2
3
4
6
7
8
9
y
1
-2
-3
-5
-4
-4
-2
3
An approximate model is y = 0.430x 2 - 4.19x + 4.95.
Comparisons will vary.
3
2
1
0
-1
-2
-3
-4
-5
1 2 3 4 5 6 7 8 9
f(x)
x
1 2 3 4 5 6 7 8 9 10
c
25
30
35
40
45
50
55
60
p
2
17
28
30
34
26
21
7
36
32
28
24
20
16
12
8
4
0
© Houghton Mifflin Harcourt Publishing Company · Image Credits: ©Gunter
Marx/Alamy
A town is holding its annual music festival 6 months from
now. Based on last year’s ticket sales and survey results of how
many tickets could be sold at different prices, a study group
presents the table and graph shown. The variable c represents
the cost of a ticket (in dollars), and P represents the estimated
overall profit (in thousands of dollars) the town would make at
that ticket price.
Profit (thousands of $)
7.
x
0
Comparisons will vary.
6.
f(x)
9
8
7
6
5
4
3
2
1
P
c
5 10 15 20 25 30 35 40 45 50 55 60
Ticket price ($)
Module 3
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164
Lesson 3
1/10/15 6:42 PM
Fitting Quadratic Functions to Data
164
a. Find an approximate quadratic function model for data by
estimating the coordinates of the vertex and using one other
point. Then sketch your graph.
COMMUNICATING MATH
Students should be able to explain how they
approximated a vertex or x–intercepts from a graph
of the data. They should also be able to address the
level of certainty to which they believe their
approximations are reasonably accurate.
Sample answer, using (45, 34) as an estimate for the
vertex and the point (25, 2):
a(x - h) + k
2
a(x - 45) + 34
2
a(25 - 45) + 34
2
a(-20)
2
-32 = a
_
400
a = -0.08
QUESTIONING STRATEGIES
f(x) = -0.08(x - 45) + 34.
What characteristics of the data would lead
you to use intercept form instead of vertex
form when fitting a quadratic function to a set of
data? If, after plotting the data points, it is easier to
approximate the x–intercepts than it is to
approximate the vertex of the model, then I would
use intercept form.
2
36
32
28
24
20
16
12
8
4
P
c
0
An approximate model is P = -0.08(c - 45) + 34.
2
5 10 15 20 25 30 35 40 45 50 55 60
Ticket price ($)
b. Find an approximate quadratic function model for the graphed
data by estimating the x-intercepts and using one other point.
Then sketch your graph.
f(x)
f(x)
28
28
a(x - x 1)(x - x 2)
a(x - 24)(x - 62)
a(35 - 24)(35 - 62)
a(11)(-27)
28
_
= a
-297
a ≈ -0.0943
f(x) = -0.0943(x - 24)(x - 62).
36
32
28
24
20
16
12
8
4
P
c
0
5 10 15 20 25 30 35 40 45 50 55 60
Ticket price ($)
An approximate model is P = -0.094(c - 24)(c - 62).
c. Use the quadratic regression feature on a graphing
calculator to find an approximate quadratic function
model for the graphed data. Then sketch your graph.
Regression equation:
y = -0.0907 x 2 + 7.84x - 136. 7;
An approximate model is
P = -0.0907 c 2 + 7.84c - 137.
Profit (thousands of $)
© Houghton Mifflin Harcourt Publishing Company
=
=
=
=
Profit (thousands of $)
Sample answer, using 24 and 62 as estimates for the
x-intercepts and the point (35, 28):
AVOID COMMON ERRORS
When solving real-world problems involving
quadratic regressions, students may not know which
variable to use as the independent variable and which
to use as the dependent variable. Point out that
variables following phrases such as based upon and
with respect to usually represent the independent
variable.
=
=
=
=
Profit (thousands of $)
f(x)
f(x)
2
-32
36
32
28
24
20
16
12
8
4
0
P
c
5 10 15 20 25 30 35 40 45 50 55 60
Ticket price ($)
Module 3
A2_MTXESE353930_U2M03L3.indd 165
165
Lesson 3.3
165
Lesson 3
2/22/14 6:22 AM
d. What do the x-intercepts of the models represent?
INTEGRATE TECHNOLOGY
The x-intercepts represent the ticket prices above which and below
which ticket sales for the festival will not be profitable.
Encourage students to explore how changes to a data
set (such as omitting one or more outliers, or
including more data points) affect the regression
equation. Students can use their graphs to compare
and discuss models that fit a data set well, and those
that don’t.
e. What do the models predict for the ticket cost (to the nearest
dollar) that the town should choose to maximize profit? What
is this profit (to the nearest thousand dollars)?
using the vertex form in part a: $45; $34,000;
using the intercept form in part b:
24 + 62
ticket price: _ = $43;
2
profit: P(43) = -0.094(43 - 24)(43 - 62) ≈ 34, or about $34, 000;
using the graph of the regression model in part c: $43, $32,000
8.
Consider the following data set.
x
10
8
13
9
11
14
6
4
12
7
5
f(x)
9.14
8.14
8.74
8.77
9.29
8.1
6.13
3.1
9.13
7.26
4.74
a. Use a graphing calculator to create a scatter plot.
See part b for plot.
b. Find a quadratic regression model for the data,
and add its graph to the scatter plot.
y ≈ 0.127x 2 + 2.785x - 6.013
c. Trace along the model to the right (or use the Table feature).
To the nearest tenth, what does the model predict for the
rightmost x-intercept?
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Fitting Quadratic Functions to Data
166
JOURNAL
H.O.T. Focus on Higher Order Thinking
9.
Have students describe how to use the attributes of a
quadratic function to fit a function to a set of
quadratic data without the use of technology.
Multi-Step Problem A school is making digital backups of 16 mm educational film
reels in its library’s archives. The table shows approximate run times of the films for a
given diameter of film on the reel.
a. Use the quadratic regression feature on a graphing
calculator to find a model for the run time T as a
function of diameter d.
T ≈ 0.401d 2 + 0.792d - 4.93
b. What does the model predict for the run time of a film
reel with a diameter of 15 inches?
T(15) = 0.401(15) 2 + 0.792(15) - 4.93 ≈ 97;
The model predicts a run time of about 97 minutes.
c. Use the calculator to graph the data points and the
regression model. Visually make a rough estimate of the
x-intercept. What does it mean in the context? What does
it indicate about a reasonable domain for the model?
Film Run Times (16 mm)
Diameter (in.)
Run Time (min)
3
2.25
5
8
7
15.75
8
31.5
12
63
14
84
About 2 inches; for a diameter of two inches, the run time is 0. This makes
sense because the hub, or center part, of the reel is the frame around
which the film is wound, so there is no film to play once you get to the
hub. The reasonable domain starts at the hub diameter, which the model
predicts is about 2 inches.
d. On the same screen, graph the horizontal line y = 60. Use the calculator to find
the intersection of this line with the model. What prediction by the model does
the point of intersection indicate?
© Houghton Mifflin Harcourt Publishing Company
The model predicts that for a run time of 1 hour, the diameter should be
about 11.8 inches.
10. Communicate Mathematical Ideas A student entered a data set into a graphing
calculator and had it perform a quadratic regression. She noticed that the coefficient a
in the model was very close to 0. Why might it be a good idea for her to plot the data
and examine it to see if a different model might be more appropriate?
When a = 0 in y = ax 2 + bx + c, you are left with y = bx + c, which is the
equation of a line. She should observe a plot of the data to see whether a
linear model is more appropriate.
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Lesson Performance Task
QUESTIONING STRATEGIES
How well does the regression equation model
the data for an entire calendar year? How do
you know? Poorly; for a quadratic function to
model the data before August 15 and after
November 15, the area would need to be negative.
Every year a hole forms in the ozone layer over the Antarctic as temperatures plummet during
the winter months in the Southern Hemisphere. As temperatures warm in the spring, the ozone
levels return to normal and the hole closes. The table provides data on the average size of the
ozone hole for a series of key dates during the Antarctic winter and spring.
Calendar Date
Day Number
Area of the Ozone Hole (million km 2)
Aug 15
0
4
Sept 1
17
18
Sept 15
31
23
Oct 1
47
24
Oct 15
61
22
Nov 1
78
17
Nov 15
92
10
Why does the graph not pass through the
origin? What does this indicate about the
situation? It does not pass through the origin
because on day number 0, the area of the hole was
not 0 km 2, it was 4,000,000 km 2.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Reasoning
a. Find a quadratic function that roughly models the changes in
area. You can either plot ordered pairs for the day number/
area and sketch a curve or use a graphing calculator to find a
regression.
Ask students to consider how reliable they think their
models would be for making predictions about
undocumented data points. Have them justify their
reasoning using specific examples related to the given
data.
b. Based on your model, on approximately what calendar date
would you predict that the area of the ozone hole will return
to zero?
Since the parabola opens downward, a must be
negative. Students can approximate a value for
a by substituting for various values of x and
finding an average value. If students use a
graphing calculator, they should find the regression
Area (millions km2)
f(x) = a(x - 47) 2 + 24.
25
y
20
15
10
5
x
0
20
f(x) = - 0.0081x 2 + 0.80x + 5.0.
40
60
80
100
Day Number
This graph shows both the ordered pairs and the regression curve.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Patterns
© Houghton Mifflin Harcourt Publishing Company
a. If students plot ordered pairs, they should
see from their graph that the pattern of the data
roughly follows a parabola with its vertex at about
(47, 24), and so a quadratic function that models
the ozone changes would be of the form
Discuss with students how the data (and the
phenomena themselves) are quadratic in nature. Help
them to describe how quadratic data behaves, and
how it differs from data that is linear in nature or is
best modeled by an absolute value function. Use
graphs and specific contexts to help illustrate the
differences among the types of models.
b. Students will probably predict that the area will return to zero around November 28 (Day
105). You may wish to discuss with students that the ozone hole on average does not close
up (that is, return to zero area) until sometime in December. The quadratic function is only
a rough model in this case; the data are not symmetrical because the ozone hole grows
more rapidly than it shrinks.
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Have students research how the hole in the ozone layer over the Antarctic has
been changing over the past 25 years. Have them compile data, similar to that in
the table above, for several different years during this time period. Instruct them
to fit a function to each set of data, and create graphs of their functions so that
they can compare the data sets.
Have students use their graphs to validate or disprove trends described by
scientists, environmentalists, and politicians. Note that reputable sources are
crucial when researching controversial topics.
1/10/15 6:42 PM
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.
Fitting Quadratic Functions to Data
168