correctionkey=nl-a;ca-a correctionkey=nl-c;ca …1, 2) (-3, (0, 1) 1 8) (-1, 1 2) (- 2, 1 4) f(x )...

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Explore 1 Graphing and Analyzing f (x) = 2 x and f (x) = 10 x

An exponential function is a function of the form ƒ (x) = b x , where the base b is a positive constant other than 1 and the exponent x is a variable. Notice that there is no single parent exponential function because each choice of the base b determines a different function.

A Complete the input-output table for each of the parent exponential functions below.

x f (x) = 2 x x p (x) = 10 x

-3 -3

-2 -2

-1 -1

0 0

1 1

2 2

3 3

B Graph the parent functions ƒ (x) = 2 x and p (x) = 10 x by plotting points.

Resource Locker

420-4 -2

8

6

4

2

x

y(3, 8)

(2, 4)

(1, 2)(0, 1)(-3, 1

8)

(-1, 12

)

(-2, 14

)

f(x)

420-4 -2

8

6

4

2

x

y

(0, 1)(-3, 11000

)

(-2, 1100

)

(-1, 110

)

p(x)

1 _ 8

1 _ 1000

1 _ 100

1 _ 10

1 _ 4

1 _ 2

1

2

4

8

1

10

100

1000

Module 13 635 Lesson 1

13.1 Exponential Growth FunctionsEssential Question: How is the graph of g (x) = a b x - h + k where b > 1 related to the graph

of f (x) = b x ?

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A2_MNLESE385900_U6M13L1.indd 635 3/22/14 11:29 AM

Common Core Math StandardsThe student is expected to:

F-BF.3

Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Also F-IF.7e, F-LE.2, A-REI.11

Mathematical Practices

MP.4 Modeling

Language ObjectiveIn a small group, match graphs to their corresponding exponential growth functions.

HARDCOVER PAGES 463472

Turn to these pages to find this lesson in the hardcover student edition.

Exponential Growth Functions

ENGAGE Essential Question: How is the graph of g (x) = a b x - h + k where b >1 related to the graph of f (x) = b x ?The graph of g (x) = a b x - h + k involves transformations of the graph of f (x) = b x . In particular, the graph of g (x) is a vertical stretch or compression of the graph of f (x) by a factor of |a|, a reflection of the graph across the x-axis if a < 0, and a translation of the graph h units horizontally and k units vertically.

PREVIEW: LESSON PERFORMANCE TASKView the Engage section online. Discuss the photo and the possible ways to model the growth of an investment over time. Then preview the Lesson Performance Task.

635

HARDCOVER

Turn to these pages to find this lesson in the hardcover student edition.

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Explore 1 Graphing and Analyzing f (x) = 2 x

and f (x) = 10 x

An exponential function is a function of the form ƒ (x) = b x , where the base b is a positive constant other than 1 and

the exponent x is a variable. Notice that there is no single parent exponential function because each choice of the base

b determines a different function.

Complete the input-output table for each of the parent exponential functions below.

xf (x) = 2 x

x p (x) = 10 x

-3

-3

-2

-2

-1

-1

0

0

1

1

2

2

3

3

Graph the parent functions ƒ (x) = 2 x and p (x) = 10 x by plotting points.

Resource

Locker

F.BF.3 For the full text of this standard, see the table starting on page CA2 of Volume 1. Also

F.IF.7e, F.LE.2, A.REI.11

420-4 -2

8

6

4

2x

y(3, 8)

(2, 4)

(1, 2)

(0, 1)(-3,

18

)

(-1, 12

)

(-2, 14

)

f(x)

420-4 -2

8

6

4

2x

y

(0, 1)(-3,

11000

)

(-2, 1

100)

(-1, 110

)

p(x)

1 _ 8

1 _

1000

1 _

100

1 _ 10

1 _ 4

1 _ 2

1

2

4

8

1

10

100

1000

Module 13

635

Lesson 1

13 . 1 Exponential Growth Functions

Essential Question: How is the graph of g (x) = a b x - h + k where b > 1 related to the graph

of f (x) = b x ?

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CorrectionKey=NL-A;CA-A

A2_MNLESE385900_U6M13L1.indd 635

3/22/14 11:27 AM

635 Lesson 13 . 1

L E S S O N 13 . 1

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C What is the domain of each function?

Domain of ƒ (x) = 2 x : ⎧

⎨ ⎩ x|

⎫

⎬ ⎭

Domain of p (x) = 10 x : ⎧

⎨ ⎩ x|

⎫

⎬ ⎭

D What is the range of each function?

Range of ƒ (x) = 2 x : ⎧

⎨ ⎩ y|

⎫

⎬ ⎭

Range of p (x) = 10 x : ⎧

⎨ ⎩ y| ⎫

⎬ ⎭

E What is the y-intercept of each function?

y-intercept of ƒ (x) = 2 x :

y-intercept of p (x) = 10 x :

F What is the trend of each function?

In both ƒ (x) = 2 x and p (x) = 10 x , as the value of x increases, the value of y increases/decreases.

Reflect

1. Will the domain be the same for every exponential function? Why or why not?

2. Will the range be the same for every exponential function in the form ƒ(x) = b x , where b is a positive constant? Why or why not?

3. Will the value of the y-intercept be the same for every exponential function? Why or why not?

Explain 1 Graphing Combined Transformations of f (x) = b x Where b > 1

A given exponential function g (x) = a ( b x - h ) + k with base b can be graphed by recognizing the differences between the given function and its parent function, ƒ (x) = b x . These differences define the parameters of the transformation, where k represents the vertical translation, h is the horizontal translation, and a represents either the vertical stretch or compression of the exponential function and whether it is reflected across the x-axis.

You can use the parameters in g (x) = a ( b x - h) + k to see what happens to two reference points during a transformation. Two points that are easily visualized on the parent exponential function are (0, 1) and (1, b) .

In a transformation, the point (0, 1) becomes (h, a + k) and (1, b) becomes (1 + h, ab + k) . The asymptote y = 0 for the parent function becomes y = k.

- ∞ < x < ∞

- ∞ < x < ∞

0 < x < ∞

0 < x < ∞

1

1

Yes, because the value of the exponent x can be any real number.

Yes, because the value of b x where b is a positive number greater than 1 will always be a

positive number.

Yes, because a non-zero based raise to the power of 0 is always equal to 1.


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A2_MNLESE385900_U6M13L1 636 21/08/14 3:44 PM

EXPLORE Graphing and Analyzing f (x) = 2 x and f (x) = 10 x

QUESTIONING STRATEGIESDoes the graph of f (x) = b x have a vertical asymptote? Why or why not? No; the

function is defined for all real values of x.

Why can’t the value of b be 1? When b = 1, the function is the constant function f (x) = 1.

EXPLAIN 1 Graphing Combined Transformations of f (x) = b x Where b > 1

INTEGRATE MATHEMATICAL PRACTICESFocus on ModelingMP.4 Visual learners can use colored pencils to sketch each of the intermediate steps in obtaining a graph of g (x) , a combined transformation from the graph of f (x) . For example, students might first draw the horizontal translation, then a reflection across the x-axis, and finally the vertical translation.

PROFESSIONAL DEVELOPMENT

Learning ProgressionsIn this lesson, students see that an exponential function is not just a function with an exponent, but a function whose independent variable is in the exponent. Therefore, f (x) = 2 x is an exponential function, but g (x) = x 2 is not. Students will graph most of the exponential functions in this lesson by hand and will see that the graphs of exponential growth functions approach a horizontal asymptote as x decreases without bound. In the next lesson, they will examine exponential decay functions, which approach the horizontal asymptote as x increases without bound.

Exponential Growth Functions 636


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The graphs of ƒ (x) = 2 x and p (x) = 1 0 x are shown below with the reference points and asymptotes labeled.

f (x) = 2 x p (x) = 10 x

Example 1 State the domain and range of the given function. Then identify the new values of the reference points and the asymptote. Use these values to graph the function.

g (x) = -3 ( 2 x - 2 ) + 1

The domain of g (x) = - 3 ( 2 x - 2 ) + 1 is ⎧ ⎨ ⎩ x| - ∞ < x < ∞ ⎫

⎬ ⎭ .

The range of g (x) = - 3 ( 2 x - 2 ) + 1 is ⎧ ⎨ ⎩ y|y < 1 ⎫

⎬ ⎭ .

Examine g (x) and identify the parameters.

a = - 3, which means that the function is reflected across the x-axis and vertically stretched by a factor of 3.

h = 2, so the function is translated 2 units to the right.

k = 1, so the function is translated 1 unit up.

The point (0, 1) becomes (h, a + k) . (h, a + k) = (2, - 3 + 1)

= (2, - 2)

(1, b) becomes (1 + h, ab + k) . (1 + h, ab + k) = (1 + 2, - 3 (2) + 1) = (3, - 6 + 1)

= (3, - 5)

The asymptote becomes y = k.

y = k → y = 1

Plot the transformed points and asymptote and draw the curve.

y = 0

y

0-2-4

2

4

6

8

42

x

f(x)

(0, 1)(1, 2)

y

0-2-4

2

4

6

8

42

x

p(x)

(0, 1) y = 0

f(x)

y = 1

(2, -2)

(3, -5)

y

0-2-4

2

-2

-4

-6

42

x



A2_MNLESE385900_U6M13L1 637 10/06/15 2:17 AM

COLLABORATIVE LEARNING

Peer-to-Peer ActivityHave pairs of students work together to make a table showing the effects of the constant a on the graphs of g (x) = a b x - h + k if a > 1 and if a < -1. In the table, have students include a sketch of each graph, as well as information about how changing the value of a affects function characteristics such as asymptotes and end behavior.

AVOID COMMON ERRORSWhen translating to obtain the graph of f (x) = b x - h , students often interpret the sign of h incorrectly and translate the parent graph in the wrong direction. A major cause of this error is the presence of the subtraction sign in the expression x - h. Remind students that this subtraction sign is not the sign of h.

637 Lesson 13 . 1


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B q (x) = 1.5 ( 10 x - 3 ) - 5

The domain of q (x) = 1.5 (1 0 x - 3 ) - 5 is ⎧

⎨ ⎩ x|

⎫

⎬ ⎭ .

The range of q (x) = 1.5 (1 0 x - 3 ) - 5 is ⎧

⎨ ⎩ y|

⎫

⎬ ⎭ .

Examine q (x) and identify the parameters.

a = so the function is stretched vertically by a factor of 1.5.

h = so the function is translated 3 units to the right.

k = so the function is translated 5 units down.

The point (0,1) becomes (h, a + k) . (h, a + k) = (3, 1.5 - 5) =

(1, b) becomes (1 + h, ab + k) . (1 + h, ab + k) = (1 + 3, 1.5 (10) - 5) =

The asymptote becomes y = k.

y = k → y =

Plot the transformed points and asymptote and draw the curve.

Your Turn

4. g (x) = 4 ( 2 x + 2 ) - 6

5. q (x) = - 3 _ 5 (1 0 x + 2 ) + 3

y

0-4-8-12

4

8

12

-44

x

q(x)

(4, 10)

(3, -3.5)

y = -5

y

0-2-4-6-8

2

-2

-4

-6

x

y = -6

(-2, -2)

(-1, 2)

y

0-2 2-4-6

2

-2

-4

x

y = 3

(-1, -3)

(-2, )125 q(x)

- ∞ < x < ∞

y > - 5

- 5

(4, 10)

(3, - 3.5)

1.5

- 5

3

The domain of g (x) = 4 ( 2 x + 2 ) - 6 is ⎧ ⎨ ⎩ x| - ∞ < x < ∞ ⎫

⎬ ⎭ .

The range of g (x) = 4 ( 2 x + 2 ) - 6 is ⎧ ⎨ ⎩ y|y > - 6

⎫ ⎬ ⎭ .

a = 4 h = - 2 k = - 6

The asymptote becomes y = -6.

The domain of q (x) = - 3 _ 5

(1 0 x + 2 ) + 3 is ⎧ ⎨ ⎩ x| - ∞ < x < ∞ ⎫

⎬ ⎭ .

The range of q (x) = - 3 _ 5

(1 0 x + 2 ) + 3 is ⎧ ⎨ ⎩ y|y < 3

⎫ ⎬ ⎭ .

a = - 3 _ 5

h = -2 k = 3

The asymptote becomes y = 3.




A2_MNLESE385900_U6M13L1.indd 638 3/31/14 9:21 PM

QUESTIONING STRATEGIESCan you transform the graph of f (x) = b x so that it has a turning point? No; the graph will

be ever increasing from the horizontal asymptote or ever decreasing to the horizontal asymptote, no matter which transformation is used.

DIFFERENTIATE INSTRUCTION

Kinesthetic ExperienceFor kinesthetic learners, draw a coordinate plane on the board so that the origin is close to a given student’s shoulder height. Have the student stand in front of the plane, aligned with the y-axis, and then arrange his or her arms to roughly model the graph of an exponential function f (x) = ab x for different values of b for a = 1 and for a = -1.



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Explain 2 Writing Equations for Combined Transformations of f(x) = b x Where b > 1

Given the graph of an exponential function, you can use your knowledge of the transformation parameters to write the function rule for the graph. Recall that the asymptote will give the value of k and the x-coordinate of the first reference point is h. Then let y 1 be the y-coordinate of the first point and solve the equation y 1 = a + k for a.

Finally, use a, h, and k to write the function in the form g (x) = a ( b x-h) + k.

Example 2 Write the exponential function that will produce the given graph, using the specified value of b. Verify that the second reference point is on the graph of the function. Then state the domain and range of the function in set notation.

Let b = 2.

The asymptote is y = 1, showing that k = 1.

The first reference point is (- 1 _ 3 , - 1 _ 3 ) . This shows that h = - 1 _ 3 and that a + k = - 1 _ 3 .

Substitute k = 1 and solve for a.

a + k = - 1 _ 3

a + 1 = - 1 _ 3

a = - 4 _ 3

h = - 1 _ 3

k = 1

Substitute these values into g (x) = a ( b x-h ) + k to find g (x) .

g (x) = a ( b x-h ) + k

= - 4 _ 3 ( 2 x+ 1 _ 3 ) + 1

Verify that g ( 2 _ 3 ) = - 5 _ 3 .

g ( 2 _ 3 ) = - 4 _ 3 ( 2 2 _ 3 + 1 _ 3 ) + 1

= - 4 _ 3 ( 2 1 ) + 1

= - 4 _ 3 (2) + 1

= 3 _ 3 - 8 _ 3

= - 5 _ 3

The domain of g (x) is ⎧

⎨

⎩ x| -∞ < x < +∞

⎫

⎬ ⎭ .

The range of g (x) is ⎧

⎨

⎩ y|y < 1

⎫

⎬ ⎭ .

y

0-1

1

-1

1

x

( , )13

13

- -

-( , )53

23

g(x)

y = 1


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EXPLAIN 2 Writing Equations for Combined Transformations of f (x) = b x Where b > 1

INTEGRATE TECHNOLOGYStudents can check the equations they write by graphing the transformed functions on their graphing calculators. Have them use the TRACE or TABLE feature to identify coordinates of points in the resulting graph.

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B Let b = 10.

The asymptote is y = , showing that k = .

The first reference point is (-4, 4.4) . This shows that h = and

that a + k = . Substitute for k and solve for a.

a + k =

a + =

a = h = k =

Substitute these values into q (x) = a ( b x - h ) + k to find q (x) .

q (x) = a ( b x - h ) + k = ( 10 x - ) +

Verify that q (-3) = -10.

q (-3) = ( 10 -3 - ) +

= ( 10 ) +

= +

=

The domain of q (x) is . The range of q (x) is .

Your Turn

Write the exponential function that will produce the given graph, using the specified value of b. Verify that the second reference point is on the graph of the function. Then state the domain and range of the function in set notation.

6. b = 2

y

0-4-8-12

4

-4

-8

x(-4, 4.4)

(-3, -10)

y = 6

q(x)

⎧ ⎨ ⎩ x| - ∞< x < +∞ ⎫

⎬ ⎭

⎧ ⎨ ⎩ y|y < 6

⎫ ⎬ ⎭

6 6

4.4

6

6

6

1

-4

-4 6

6

6

-4-1.6

-1.6

-1.6

-1.6

-16

-10

4.4

4.4

-4

43210

1

-1

-2

x

y

g(x)(3, -1)

(4, 12)

y = - 52

The asymptote is y = - 5 _ 2 → k = - 5 _ 2 .

The first reference point is (3, -1) → h = 3 and

a + k = -1. Substitute k and solve for a.

a + k = -1

a - 5 _ 2 = -1

a = 3 _ 2

a = 3 _ 2 ; h = 3; k = - 5 _ 2

g (x) = 3 _ 2 ( 2 x - 3 ) - 5 _ 2

Verify that g (4) = 1 _ 2 .

g (4) = 3 _ 2 ( 2 4 - 3 ) - 5 _ 2 = 1 _ 2

The domain of g (x) is {x⎜-∞ < x < +∞}.

The range of g (x) is ⎧ ⎨ ⎩ y∣y > - 5 _ 2

⎫ ⎬ ⎭ .




A2_MNLESE385900_U6M13L1 640 8/20/14 1:21 PM

QUESTIONING STRATEGIESHow do you find h and k from the first reference point? The x-coordinate of the first

reference point is h. Substitute the value of k in the equation y = a + k to find a.

How do you use the second reference point? After you substitute a, h, and k and

find the function, substitute the coordinates of the second reference point to check that the function includes this point.



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Explain 3 Modeling with Exponential Growth FunctionsAn exponential growth function has the form ƒ(t) = a (1 + r)t where a > 0 and r is a constant percent increase (expressed as a decimal) for each unit increase in time t. That is, since ƒ(t + 1) = (1 + r) · ƒ(t) = ƒ(t) + r · ƒ(t), the value of the function increases by r · ƒ(t) on the interval [t, t + 1]. The base 1 + r of an exponential growth function is called the growth factor, and the constant percent increase r, in decimal form, is called the growth rate.

Example 3 Find the function that corresponds with the given situation. Then use the graph of the function to make a prediction.

Tony purchased a rare guitar in 2000 for $12,000. Experts estimate that its value will increase by 14% per year. Use a graph to find the number of years it will take for the value of the guitar to be $60,000.

Write a function to model the growth in value for the guitar.

ƒ (t) = a (1 + r) t

= 12, 000 (1 + 0.14) t

= 12, 000 (1.14) t

Use a graphing calculator to graph the function.

Use the graph to predict when the guitar will be worth $60,000.

Use the TRACE feature to find the t-value where ƒ (t) ≈ 60, 000.

So, the guitar will be worth $60,000 approximately 12.29 years after it was purchased.

At the same time that Tony bought the $12,000 guitar, he also considered buying another rare guitar for $15,000. Experts estimated that this guitar would increase in value by 9% per year. Determine after how many years the two guitars will be worth the same amount.

Write a function to model the growth in value for the second guitar.

g (t) = a (1 + r) t

= (1 + ) t

= ( ) t

Use a graphing calculator to graph the two functions.

Use the graph to predict when the two guitars will be worth the same amount.

Use the intersection feature to find the t-value where g (t) = .

So, the two guitars will be worth the same amount years after 2000.

0.09

1.09

4.98

15, 000

15, 000

f (t)




EXPLAIN 3 Modeling With Exponential Growth Functions

AVOID COMMON ERRORSStudents may multiply the initial amount by the growth factor and then raise that product to the exponent. Remind students to apply the order of operations. The exponent applies only to the growth factor.

QUESTIONING STRATEGIESWhat is the exponential growth function f (t) = a (1 + r) t if the growth rate is 0.05%?

Why? f (t) = a (1.0005) t , because 0.05% written as a decimal is 0.0005

When should you round exponential growth calculations? Round during the last step

only. Rounding earlier will compound errors.

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Reflect

7. In part A, find the average rates of change over the intervals (0, 4) , (4, 8) , and (8, 12) . Do the rates increase, decrease, or stay the same?

Your Turn

Find the function that corresponds with the given situation. Then graph the function on a calculator and use the graph to make a prediction.

8. John researches a baseball card and finds that it is currently worth $3.25. However, it is supposed to increase in value 11% per year. In how many years will the card be worth $26?

Elaborate

9. How are reference points helpful when graphing transformations of ƒ (x) = b x or when writing equations for transformed graphs?

10. Give the general form of an exponential growth function and describe its parameters.

11. Essential Question Check-In Which transformations of f (x) = b x change the function’s end behavior? Which transformations change the function’s y-intercept?

The average rate of change from 0 to 4 is 20, 267.52 - 12, 000 ________________ 4 - 0 = 2066.88, the average rate of

change from 4 to 8 is 34, 231.04 - 20, 267.52 _______________ 8 - 4 = 3490.88, and the average rate of change from

8 to 12 is 57, 814.86 - 34, 231.04 _______________ 12 - 8 = 5895.96. The average rates of change increase from one

interval to the next, so the rates are increasing for increasing values of x.

Write a function to model the growth in value for the baseball card.

g (t) = a (1 + r) t = 3.25 (1 + 0.11) t = 3.25 (1.11) t .

Graphing the function and using the trace function to determine when g(x) ≈ 26 result in x ≈ 20.

So, the card will be worth $26 approximately 20 years after John purchased it.

Reference points help when graphing transformations because all of the points of the

parent graph are affected in the same manner. So by shifting each reference point by

the same value or ratio, the transformed graph can be drawn. Similarly, the points of

a transformed graph can be compared to the reference points on the parent graph to

determine the values of a, h, and k in order to find the equation of the transformed graph.

A function f (x) = ab x is an exponential growth function, where a > 0 and b > 1.

Reflections across the x-axis change the function’s end behavior by making the reflected

function increase toward the asymptote as x tends to negative infinity and decrease

toward negative infinity as x tends to infinity. Vertical translations also change the

function’s end behavior since they affect the graph’s asymptote. Vertical translations,

horizontal translations, vertical stretches and compressions, and reflections across the

x-axis all affect the location of the graph’s y-intercept.




A2_MNLESE385900_U6M13L1 642 10/17/14 1:50 AM

ELABORATE CONNECT VOCABULARY Have students begin to create a list of words associated with exponential functions, and later their inverses, beginning with growth and interest, then later adding vocabulary such as decay, e, and logarithms.

INTEGRATE MATHEMATICAL PRACTICESFocus on Critical ThinkingMP.3 When students have the option of several points to use as a reference point and the y-intercept is one such point, students should find that it is easier to use the y-intercept than the other points for calculations.

SUMMARIZE THE LESSONWhat are the effects of the constant a on the graph of g (x) = a b x - h + k? If a > 1, the

graph of the parent function f (x) = b x is stretched vertically by a factor of a. If 0 < a < 1, the graph is shrunk vertically by a factor of a. If a < -1, the graph of f (x) = b x is stretched vertically by a factor of |a| and reflected across the x-axis. If -1 < a < 0, the graph of f (x) = b x is shrunk vertically by a factor of |a| and reflected across the x-axis.

LANGUAGE SUPPORT

Communicate MathHave students work in pairs. Each student writes an exponential growth function in the form f (x) = b x on an index card, and sketches its graph on another card. Students shuffle the index cards and lay them face down in an array. They take turns turning two cards over, trying to match the graph of the function to its equation. Students must justify why they think they do or don’t have a match.



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Describe the effect of each transformation on the parent function. Graph the parent function and its transformation. Then determine the domain, range, and y-intercept of each function.

1. ƒ (x) = 2 x and g (x) = 2 ( 2 x ) 2. ƒ (x) = 2 x and g (x) = -5 ( 2 x )

3. ƒ (x) = 2 x and g (x) = 2 x + 2 4. ƒ (x) = 2 x and g (x) = 2 x + 5

• Online Homework• Hints and Help• Extra Practice

Evaluate: Homework and Practice

840-8 -4

8

-8

4

-4

x

y

f(x)

g(x)

840-8 -4

8

-8

4

-4

x

y

f(x)

g(x)

840-8 -4

8

-8

4

-4

x

y

f(x)

g(x)

840-8 -4

8

-8

4

-4

x

yf(x)g(x)

g (x) is a vertical stretch (or horizontal compression) of f (x) . The domain of f (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The domain of g (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The y-intercept of f (x) is 1 and the y-intercept of g (x) is 2.

g (x) is a vertical stretch (or horizontal compression) of f (x) combined with a reflection about the x-axis. The domain of f (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The domain of g (x) is {x∣-∞ < x < ∞} and the range is {y∣y < 0}. The y-intercept of f (x) is 1 and the y-intercept of g (x) is -5.

g (x) is a shift to the left by 2 of f (x) . The domain of f (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The domain of g (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The y-intercept of f (x) is 1 and the y-intercept of g (x) is 4.

g (x) is a shift upward by 5 of f (x) . The domain of f (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The domain of g (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 5}. The y-intercept of f (x) is 1 and the y-intercept of g (x) is 6.



A2_MNLESE385900_U6M13L1.indd 643 3/31/14 9:27 PMExercise Depth of Knowledge (D.O.K.) Mathematical Practices

1–8 1 Recall of Information MP.6 Precision

9–14 2 Skills/Concepts MP.2 Reasoning

15–17 2 Skills/Concepts MP.2 Reasoning

18–22 2 Skills/Concepts MP.4 Modeling

23 3 Strategic Thinking MP.2 Reasoning

24 3 Strategic Thinking MP.3 Logic

EVALUATE

ASSIGNMENT GUIDE

Concepts and Skills Practice

ExploreGraphing and Analyzing f (x) = 2 x and f (x) = 10 x

Example 1Graphing Combined Transformations of f (x) = b x where b > 1

Exercises 9–14

Example 2Writing Equations for Combined Transformations of f (x) = b x where b > 1

Exercises 15–17

Example 3Modeling With Exponential Growth Functions

Exercises 18–22

AVOID COMMON ERRORSWhen translating the graph of g (x) = a b x - h + k, students commonly interpret the sign of h incorrectly and translate the parent graph in the wrong direction. A major cause of this error is the presence of the subtraction sign in the expression x - h. Remind students that this subtraction sign is not the sign of h. In the expression x - 2, for instance, h = 2 (a positive number), so the translation is to the right. To identify the value of h in the expression x + 2, students should rewrite it as x - (-2) , which means that h = -2, so the translation is to the left.

643 Lesson 13 . 1


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5. ƒ (x) = 10 x and g (x) = 2 (1 0 x ) 6. ƒ (x) = 10 x and g (x) = -4 (1 0 x )

7. ƒ (x) = 10 x and g (x) = 1 0 x - 2 8. ƒ (x) = 10 x and g (x) = 1 0 x - 6

9. Describe the graph of g (x) = 2 (x - 3) - 4 in terms of ƒ (x) = 2 x .

10. Describe the graph of g (x) = 1 0 (x + 7) + 6 in terms of ƒ (x) = 10 x .

420-4 -2

8

6

2

4

x

y

f(x)

g(x)

840-8 -4

8

-8

4

-4

x

y

f(x)

g(x)

420-4 -2

8

6

2

4

x

y

f(x)

g(x)840-8 -4

8

-8

4

-4

x

y

f(x)

g(x)

g (x) is a vertical stretch (or horizontal compression) of f (x) . The domain of f (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The domain of g (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The y-intercept of f (x) is 1 and the y-intercept of g (x) is 2.

g (x) is a vertical stretch (or horizontal compression) of f (x) combined with a reflection about the x-axis. The domain of f (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The domain of g (x) is {x∣-∞ < x < ∞} and the range is {y∣y < 0}. The y-intercept of f (x) is 1 and the y-intercept of g (x) is -4.

g (x) is a shift to the right by 2 of f (x) . The

domain of f (x) is {x∣-∞ < x < ∞} and

the range is {y∣y > 0}. The domain of

g (x) is {x∣-∞ < x < ∞} and the range is

{y∣y > 0}. The y-intercept of f (x) is 1

and the y-intercept of g (x) is 1 ___ 100 .

g (x) is a shift downward by 6 of f (x) . The domain of f (x) is {x∣-∞ < x < ∞} and the range is {y∣y > 0}. The domain of g (x) is {x∣-∞ < x < ∞} and the range is {y∣y > -6}. The y-intercept of f (x) is 1 and the y-intercept of g (x) is -5.

The graph is shifted down by 4 and right by 3.

The graph is shifted up by 6 and left by 7.







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State the domain and range of the given function. Then identify the new values of the reference points and the asymptote. Use these values to graph the function.

11. h (x) = 2 ( 3 x + 2 ) - 1

12. k (x) = -0.5 ( 4 x - 1 ) + 2

13. ƒ (x) = 3 ( 6 x - 7 ) -8

2-2-4-6 0

6

-2

4

2

x

y

h(x)

(-1, 5)

(-2, 1)

y = -1

3210

2

-2

1

-1

x

y

k(x)

(1, 1.5)

(2, 0)

y = 2

840-8 -4

8

-8

4

-4

x

y

f(x)

y = -8

(7, -5)

(8, 10)

The domain of h (x) = 2 ( 3 x + 2 ) - 1 is {x∣-∞ < x < ∞}.

The range of h (x) = 2 ( 3 x + 2 ) - 1 is {y∣y > -1}.

a = 2 so the function is vertically stretched by a factor of 2.

h = -2 so the function is translated 2 units to the left.

k = -1 so the function is translated 1 unit down.

The point (0, 1) becomes (-2, 1) .



The domain of k (x) = -0.5 ( 4 x-1 ) + 2 is {x∣-∞ < x < ∞}.

The range of k (x) = -0.5 ( 4 x-1 ) + 2 is {y∣y < 2}.

a is negative so function is reflected across x–axis

a = -0.5 so the function is vertically compressed by a factor of 1 __ 2 .

h = 1 so the function is translated 1 unit to the right.

k = 2 so the function is translated 2 units up.

The point (0, 1) becomes (1, 1.5) .

The point (1, 4) becomes (2, 0) .


The domain of f (x) = 3 ( 6 x - 7 ) -8 is {x|-∞ < x < ∞}.

The range of f (x) = 3 ( 6 x - 7 ) -8 is {y|y > -8}.


h = 7 so the function is translated 7 units to the right.

k = -8 so the function is translated 8 units down.

The point (0, 1) becomes (7, -8) .





A2_MNLESE385900_U6M13L1 645 8/20/14 1:22 PM

PEER-TO-PEER DISCUSSION Ask students to discuss with a partner how they would transform f (x) = 1.1 x and g (x) = - 1.1 x . Have them choose a translation and a stretch, write the rule, and sketch the graphs. If desired, have students choose an axis to reflect the function across and repeat.

645 Lesson 13 . 1


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14. ƒ (x) = -3 ( 2 x + 1 ) +3

15. h (x) = - 1 _ 4 ( 5 x + 1 ) - 3 _ 4

16. p (x) = 2 ( 4 x - 3 ) -5

420-4 -2

4

-4

2

-2

x

y

g(x)

(0, -3)

(-1, 0)

y

0-2-4

2

4

-2

-4

42

x

(4, 3)

(3, -3)

p(x)

y

h(x)

0-2-4

2

4

-2

-4

42

x

(-1, -1) (0, -2)

The domain of f (x) = -3 ( 2 x + 1 ) +3 is {x|-∞ < x < ∞}.

The range of f (x) = -3 ( 2 x + 1 ) +3 is {y|y < 3}.


a = -3 so the function is vertically stretched by a factor of 3.

h = -1 so the function is translated 1 unit to the left.

k = 3 so the function is translated 3 units up.




The domain of h (x) = - 1 _ 4

( 5 x + 1 ) - 3 _ 4

is {x|-∞ < x < ∞}.

The range of h (x) = -1 _ 4

( 5 x + 1 ) - 3 _ 4

is ⎧ ⎨ ⎩ y|y < - 3 _

4 ⎫ ⎬ ⎭ .


a = - 1 _ 4

so the function is vertically compressed by a factor of 1 __ 4 .

h = -1 so the function is translated 1 unit to the left.

k = - 3 _ 4

so the function is translated 3 __ 4 unit down.

The point (0, 1) becomes (-1, -1) .


The asymptote becomes y = - 3 _ 4

.

The domain of p (x) = 2 ( 4 x - 3 ) -5 is {x|-∞ < x < ∞}.

The range of p (x) = 2 ( 4 x - 3 ) -5 is {y|y > -5}


h = 3 so the function is translated 3 unit to the right.

k = -5 so the function is translated 5 unit down.










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Write the exponential function that will produce the given graph, using the specified value of b. Verify that the second reference point is on the graph of the function. Then state the domain and range of the function in set notation.

17. b = 2

18. b = 10

y

0-2-4-6

2

4

8

2

x

(-1, 5.2)

(0, 7.4)

y = 3

g(x)

(-2, 3.7)

(-1, 1)

y

0-2-4-6-8

2

x

q(x)

-2

-4

4y = 4

The asymptote is y = 3 → k = 3.

The first reference point is (-1, 5.2) → h = -1

and a + k = 5.2. Substitute for k and solve for a.

a + k = 5.2

a + 3 = 5.2

a = 2.2

a = 2.2; h = -1; k = 3

Substitute these values into g (x) = a ( b x - h ) + k to find g (x) .

g (x) = 2.2 ( 2 x + 1 ) + 3

Verify that g (0) = 7.4.

g (0) = 2.2 ( 2 0 + 1 ) + 3 = 7.4

The domain of g (x) is {x|-∞ < x < + ∞}.

The range of g (x) is {y|y > 3}.

The asymptote is y = 4 → k = 4.

The first reference point is (-2, 3.7) → h = -2 and

a + k = 3.7. Substitute for k and solve for a.

a + k = 3.7

a + 4 = 3.7

a = -0.3

a = -0.3; h = -2; k = 4

Substitute these values into q (x) = a ( b x - h ) + k to find q (x) .

q (x) = -0.3 ( 10 x + 2 ) + 4

Verify that q (-1) = 1.

q (-1) = -0.3 (1 0 -1 + 2 ) + 4 = 1

The domain of q (x) is {x|-∞ < x <+∞}.

The range of q (x) is {y|y < 4}.




INTEGRATE TECHNOLOGYStudents may find it useful to use a spreadsheet to calculate values they use repeatedly. For example, after they enter the values of a, b, h, and k in separate cells, they can calculate coordinates of two reference points as formulas using these values.

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Find the function that corresponds with the given situation. Then graph the function on a calculator and use the graph to make a prediction.

19. A certain stock opens with a price of $0.59. Over the first three days, the value of the stock increases on average by 50% per day. If this trend continues, how many days will it take for the stock to be worth $6?

20. Sue has a lamp from her great-grandmother. She has it appraised and finds it is worth $1000. She wants to sell it, but the appraiser tells her that the value is appreciating by 8% per year. In how many years will the value of the lamp be $2000?

21. The population of a small town is 15,000. If the population is growing by 5% per year, how long will it take for the population to reach 25,000?

f (t) = a (1 + r) t

= 0.59 (1 + 0.5) t

= 0.59 (1.5) t

Graphing the function and using the trace function to determine

when f (x) = 6 result in x ≈ 5.74.

So, it will take approximately 5.74 days for the stock to be worth $6.

f (t) = a (1 + r) t

= 1000 (1 + 0.08) t

= 1000 (1.08) t

Graphing the function and using the trace function

to determine when f (x) = 2000 result in x ≈ 9.04.

So, it will take approximately 9.04 years for the value of the

lamp to be $2000.

f (t) = a (1 + r) t

= 15,000 (1 + 0.05) t

= 15,000 (1.05) t

Graphing the function and using the trace function to determine when

f (x) = 25,000 result in x ≈ 10.47.

So, it will take approximately 10.47 years for the population

to reach 25,000.




A2_MNLESE385900_U6M13L1 648 8/20/14 1:23 PM



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22. Bill invests $3000 in a bond fund with an interest rate of 9% per year. If Bill does not withdraw any of the money, in how many years will his bond fund be worth $5000?

H.O.T. Focus on Higher Order Thinking

23. Analyze Relationships Compare the end behavior of g (x) = 2 x and f (x) = x 2 . How are the graphs of the functions similar? How are they different?

24. Explain the Error A student has a baseball card that is worth $6.35. He looks up the appreciation rate and finds it to be 2.5% per year. He wants to find how much it will be worth after 3 years. He writes the function ƒ (t) = 6.35 (2.5) t and uses the graph of that function to find the value of the card in 3 years.

According to his graph, his card will be worth about $99.22 in 3 years. What did the student do wrong? What is the correct answer?

f (t) = a (1 + r) t

= 3000 (1 + 0.09) t

= 3000 (1.09) t

Graphing the function and using the trace function to determine when

f (x) = 5000 result in x ≈ 5.96.

So, it will take approximately 5.96 years for Bill’s bond fund to reach

$5000.

For g (x) = 2 x , as x → + ∞, g (x) → + ∞ and as x → - ∞, g (x) → 0. For f (x) = x 2 , as x → + ∞, f (x) → + ∞ and as x → - ∞, f (x) → + ∞. The two functions have the same end behavior as x → + ∞, however as x → - ∞, f (x) = x 2 is unbounded, whereas g (x) = 2 x is bounded below by 0.

The student forgot to convert the growth rate to a decimal. The function

should be f (t) = 6.35 (1.025) t . Using the correct function and graph gives

f (3) = 6.8382555. So, in three years, the card will be worth about $6.84.




JOURNALHave students describe the steps they would take to graph f (x) = 2 x + 1 - 1 and describe and explain the transformations from the parent function.

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Like all collectables, the price of an item is determined by what the buyer is willing to pay and the seller is willing to accept. The estimated value of a 1948 Tucker 48 automobile in excellent condition has risen at an approximately exponential rate from about $500,000 in December 2006 to about $1,400,000 in December 2013.

a. Find an equation in the form V (t) = V 0 (1 + r) t , where V 0 is the value of the car in dollars in December 2006, r is the average annual growth rate, t is the time in years since December 2006, and V (t) is the value of the car in dollars at time t. (Hint: Substitute the known values and solve for r.)

b. What is the meaning of the value of r?

c. If this trend continues, what would be the value of the car in December 2017?

Lesson Performance Task

a. December 2013 is 7 years from December 2006. Substitute

500,000 for V 0 , 7 for t, and 1,400,000 for V (t) .

1,400,000 = 500,000 (1 + r) 7

Solve for r.

1,400,000 _______ 500,000 =

500,000 (1 + r) 7 __________ 500,000

2.8 = (1 + r) 7

7 √ ―― 2.8 =

7 √ ―――

(1 + r) 7

1.158 ≈ 1 + r

0.158 ≈ r

V (t) = 500,000 (1.158) t

b. The average annual growth rate for the value of a 1948 Tucker 48

is about 15.8%.

c. In December 2017, t = 11.

V (11) = 500,000 (1.158) 11

≈ 2,510,522.73

The car would have a value of about $2,510,522.73.



A2_MNLESE385900_U6M13L1 650 8/20/14 1:25 PM

EXTENSION ACTIVITY

Have students research the average growth rate of the stock market from 2006 to 2013. Then have students write an exponential growth function V (t) = V 0 (1 + r) t for the average growth of an initial investment V 0 = $500, 000. Have students compare the value o f the car to the value of the stock investment in 2013.

QUESTIONING STRATEGIESIf the car value increased by a constant amount each year, how could the car’s value be

modeled? with a linear equation

Does the function V (t) have an upper limit? Mathematically, no. But in a

real-world situation, circumstances might limit the value of the car.

Scoring Rubric2 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.



correctionkey=nl-a;ca-a correctionkey=nl-c;ca …1, 2) (-3, (0, 1) 1 8) (-1, 1 2) (- 2, 1 4) f(x )...

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