MODELING WITH FUNCTIONSrusdmath.weebly.com/uploads/1/1/1/5/11156667/g9_u1...High School: Modeling...

MATH HIGH SCHOOL

MODELING WITH FUNCTIONS

ANSWERS FOR EXERCISES

Copyright © 2015 Pearson Education, Inc. 40

High School: Modeling with Functions

ANSWERSLESSON 2: MODELING RELATIONSHIPS

ANSWERS

F.IF.1 1. Graph A: Not a function

Graph B: Not a function

Graph C: Function

F.IF.1 2. Table A: Function

Table B: Not a function

F.IF.2 3. d CC( ) =π

F.IF.1 4. No, this graph is not a function. There is more than one y-value for an x-value. Note: This graph shows the equation x = 4.

F.IF.2 5. A(r) = π • r2

A(4) = π • 42

A(4) = π16

A(4) = 16π

F.IF.2 6. for A > 0

F.IF.2 7. f(x) = 2x – 7

F.IF.1 8. A The relationship between d and C is a function.

D This table represents the relationship.

d 1 2 3 4

C 1π 2π 3π 4π

Challenge Problem

F.IF.2 9. U(0) = –1

U(1) = 1

U(2) = 3

U(3) = 5

U(4) = 7

It is a linear function.

r AA

( ) =π



ANSWERSLESSON 3: DIFFERENT WAYS OF GROWING

ANSWERS

F.IF.2 1. C Exponential

F.LE.1.b 2. A

F.LE.1 F.LE.2

3.x 0 1 2 3 4 5 6

y 50 60 70 80 90 100 110

F.LE.1 F.LE.2

4. x 0 1 2 3 4 5 6

y 50 55 60.5 66.55 73.21 80.53 88.58

F.LE.3 5. Function A reaches 100 at t = 5, while Function B is only 80.53 when t = 5. Function B shows multiplicative growth, so it is an exponential function. Function A shows additive growth, so it is a linear function. Over time, an exponential function eventually exceeds a linear function, but in this case, Function A reaches 100 quicker than Function B.

F.IF.7.a 6.

89

10

10

234567

54 63210 x

f(x)



ANSWERSLESSON 3: DIFFERENT WAYS OF GROWING

F.IF.7.e 7.

89

10111213141516

12

0

34567

5 74 863210 x

f(x)

F.LE.3 8. The graphs show that the functions are equal somewhere between x = 2.5 and x = 16. The point of intersection is (2.53, 16.05).

Challenge Problem

F.LE.2 9. f(x) = x + 11

g(x) = 3x



ANSWERSLESSON 4: COMPARING GROWTH

ANSWERS

F.LE.1 1.a. Sarah has $650 in her piggy bank. She adds $50 each month. Linear

b. David has $400 in his bank. He earns 10% interest each year. Exponential

c. Cell division takes place as shown in the diagram. Each cell divides into a new cell every hour. Exponential

d. Miki collects souvenirs related to Michael Jackson. She allows herself to buy a new one every 3 months. She already has 27 items.

Linear

F.LE.2 2. f(x) = 650 + 50x, where x is time in months.

F.LE.1.a 3. C 6

F.LE.1.a 4. A

F.LE.3 5. In the exponential function y = 3x, as x increases from 0 to 2, y increases from 1 to 9

or at a rate of change of 4. In the linear function y x= 34

, as x increases from 0 to 2,

y increases from 0 to 64

, or at a rate of change of 34

.

An exponential function will also exceed a linear function in growth over time because the linear function increases at a constant rate while the exponential function is influenced by the number taken to a power.

F.LE..1a 6. D –2.5

F.IF.7 7.

150

100

8 10642 7 9531–5 –3 –1–2–4–6 x

y

50

34



ANSWERSLESSON 4: COMPARING GROWTH

F.IF.7 8.

100

200

300

400

8 10642 7 9531–1–2 x

y

(4, 80)

(5, 160)

–3

Challenge Problem

F.IF.7.a 9. a.

5

10

15

21–1–2 x

y

y = 4x

y =

4x2

b. Points of intersection: (0, 0) and (1, 4)

c. This solution uses algebra you might not have learned yet.

The point of intersection is where the two expressions are equal. 4x2 = 4x

4x2 – 4x = 0 4x(x – 1) = 0 Thus: 4x = 0 or x – 1 = 0 x = 0 or x = 1 If x = 0, then y = 4x or y = 4 • 0, y = 0. If x = 1, then y = 4x or y = 4 • 1, y = 4. The points of intersection are (0, 0) and (1, 4).



ANSWERSLESSON 5: DOMAIN AND RANGE

ANSWERS

F.IF.5 1. B 0 ≤ d ≤ 5.2

F.IF.5 2. C 0 ≤ h ≤ 14.5

F.1F.4 N.Q.1

3. A The height when the ball was released

F.IF.5 4. 0 < N(t) < 1 for t values less than –5; you cannot have a fraction of a cell.

F.IF.9 5. Here are some examples.

2

12

18

33

− = ( ) =

20 = 1

F.IF.2 6. C 18

F.IF.2 7. a. f(4) = 18 • 1.54 = 91.125

b. f ( ) • . •.

•.

− = = = =−2 18 1 5 181

1 518

12 25

822

F.IF.5 8. Domain: all real numbers

F.IF.5 9. Range: y = f(x) > 0

Challenge Problem

F.IF.5 10.Approximate g by using

5.015 35 14

1 035.14 .

..≈ ≈

Then f(x) = 5.01 • (1.03)x

f(15) = 5.01 • (1.03)15

f(15) ≈ 7.8

The population of Laos would be around 7.8 million in 2015.



ANSWERSLESSON 6: COMPARING GRAPHS OF FUNCTIONS

ANSWERS

F.IF.1 1. A

5

–10

10

–5

105 15–5–10–15 x

y

x y= 12

33 –

B

5

–10

10

–5

105 15–5–10–15 x

y

y x= log

F.IF.2 2. B y = 3

F.IF.2 3. B y = 3

F.IF.1 4. a. In Table A, y increases by a constant amount of 2 over each unit interval of x (add 2).

b. In Table B, y increases by a constant factor of 2 over each unit interval of x (multiply by 2).

F.IF.2 5. –2 –1 0 1 2

–5 –2 1 4 7

14

12

1 2 4

x

f(x)

g(x)



ANSWERSLESSON 6: COMPARING GRAPHS OF FUNCTION

F.IF.5 6. a. In the domain –2 ≤ x ≤ 2, f(x) is greater than g(x) for all values

x > 0, and g(x) is greater than f(x) for all values x < 0.

b. The range of f(x) is all real numbers. The range of g(x) is all real numbers > 0.

F.IF.5 7. a. In the domain –5 ≤ x ≤ 5, f(x) is always greater than g(x).

b. The range of f(x) is all real numbers. The range of g(x) is all real numbers < 0.

c. The functions never intersect.

F.IF.1 8. a. The domain of the function y = 3 includes all real numbers, but the range is just 3; y is equal to 3 for every value of x included in the domain.

b. x = 3 is not a function because its domain is just x = 3, and that single x-value generates a range that includes all real y-values.

Challenge Problem

–2

–3

–4

–5

2

3

4

5

6

7

1

–1–1 21–2 x

y

f(x)

g(x)

–4

–6

–8

–10

4

6

8

10

12

14

2

–2–2 42–4 x

y

f(x) = –2x + 4

g(x) = –2x + 4



ANSWERSLESSON 6: COMPARING GRAPHS OF FUNCTION

F.IF.1 9. a. The domain of the functions is all real numbers. The range is the same for both and cannot be less than 0, since the exponential term yields a smaller and smaller fraction as x becomes more and more negative.

b. Linear functions like these have the same domain and range of all real numbers since each line continues infinitely in both the positive and negative y-directions.



ANSWERSLESSON 7: THREE KINDS OF GROWTH

ANSWERS

F.LE.1.b 1.

Linear Function Cubic Function Exponential Function Quadratic Function

The number of insects in a population that doubles

each day

The area of a square with a side length of x

The speed of a car moving at constant speed,

with respect to time

The volume of a sphere with a radius of r

Type of Functions

Situations

F.IF.1 2. D exponential

F.IF.1 3. a. The domain is all real numbers.

b. The range is y > 0.

F.IF.1 4. B quadratic

F.IF.1 5. a. The domain is all real numbers.

b. The range is y ≥ 0.

Note: The formula is f(x) = x2.

F.IF.1 6. A linear

F.IF.1 7. a. Domain: 0 ≤ x ≤ 81.2 The domain is usually not negative when you are dealing with time.

b. Range: 0 ≤ y ≤ 2,030 with y being a whole number The range cannot be negative because the context is about people. Note: The formula is N(t) = 2,030 − 25t, where t is the number of years after 2008.

F.IF.1 8. C cubic



ANSWERSLESSON 7: THREE KINDS OF GROWTH

F.IF.1 9. a. B quadratic

b. When you look at the numbers, you see squares.

Challenge Problem

F.IF.7.a F.LE.2

10. a.

25

0

50

75

0.50.25Time

Dis

tanc

e

0 x

y

b. y = x2, where x = 1

20 of a second

2 sec will be 40 units. y = 402 y = 1,600 units

The distance after 2 sec is 1,600 units.



ANSWERSLESSON 8: LINEAR AND EXPONENTIAL DECAY

ANSWERS

F.IF.4 1. A The slope is a negative number.

C The graph slopes downward from left to right.

F.IF.4 2. C The graph slopes downward from left to right.

D The growth factor, a, is less than 1, and the exponent, x, is positive.

F.IF.4 N.Q.1

3. From the graph, the rock looks to be about 2,000 million, or 2 billion, years old.

F.LE.1.c 4. The formula is a function that represents exponential decay. The exponent is negative and is multiplied by 100 (the y-intercept).

F.IF.7 5. All the functions are linear. The functions with the negative constants, y = –3x and y = –3x – 6, are decreasing linear functions, while y = 3x and y = 3x – 6 are increasing linear functions. The domain of all the functions is all real numbers, and the range of all the functions is all real numbers. y = 3x and y = –3x are proportional relationships because they go through the point (0, 0). y = 3x – 6 and y = –3x – 6 both intersect the y-axis at (0, –6).

F.IF.7 6.

–5 –2 2–4 –1 31–3 4 5 x

y

–2

–1

5

4

3

2

1

y = 3xyx

= ⎛⎝⎜

⎞⎠⎟

13

Both functions are exponential. The function y = 3x has a growth factor greater than 1 and

is increasing. The function yx

= ⎛⎝⎜

⎞⎠⎟

13

has a growth factor between 0 and 1 and is decreasing. The domain of both functions is all real numbers, and the range of both functions is all real numbers greater than 0. They both intersect the y-axis at (0, 1).

–5 –2 2–4 –1 31–3 4 5 x

y

–5

–4

–3

–2

–1

5

4

3

2

1

y = 3xy = –3x

y = 3x – 6y = –3x – 6



ANSWERSLESSON 8: LINEAR AND EXPONENTIAL DECAY

F.BF.1 7. A y = –2t + 35

F.LE.5 8. D Decreasing exponential function

F.LE.5 A.SSE.3.c

9. B 3% per year

F.IF.2 10. 4,625

Challenge Problem

F.LE.5 11. 120 • 0.54 g or 7.5 g



ANSWERSLESSON 9: RECURSIVE RULES FOR SEQUENCES

ANSWERS

F.LE.2 1. 19, 23, 27

F.LE.2 2. C 39

F.IF.2 F.IF.3 F.LE.2

3. B f(1) = 3, f(n) = f(n – 1) + 4 for n > 1


4. You start with –17 and you add 3 to the last term to get the next term.

F.LE.2 5. There are nine missing terms: –8, –5, –2, 1, 4, 7, 10, 13, 16


6. f(1) = –17, f(n) = f(n – 1) + 3 for n > 1

F.LE.2 7. D Start with 40, then multiply the last term by to get the next term.

F.IF.3 F.LE.2

8. a. 40, –20, 10, –5, 2.5, –1.25, 0.625, –0.3125, 0.15625, –0.078125

b.

5

0

–5

–10

–15

–20

10

15

20

25

30

35

40

4 6 8 102 x

y

− 12



ANSWERSLESSON 9: RECURSIVE RULES FOR SEQUENCES


9.

F.IF.3 F.LE.2

10. 1, 1, 2, 4, 7, 13, 24, 44, 81, 149

Challenge Problem


11. a. n = 7; f(7) = 0.625

b. f(n) > n2 for n > 9

n n2 f(n)

1 1 1

2 4 1

3 9 2

4 16 4

5 25 7

6 36 13

7 49 24

8 64 44

10 100 149

9 81 81

f f n f n n( ) , ( ) ( )1 4012

1 1= = − − >for



ANSWERSLESSON 10: SEQUENCES

ANSWERS

F.LE.2 1.1, 4, 7, 10, 13, … Arithmetic sequence

1, 3, 9, 27, 81, … Geometric sequence

–18, –12, –6, 0, 6, … Arithmetic sequence

–63, –52, –41, –30, –19, … Arithmetic sequence

–63, 21, –7, , , … Geometric sequence

F.LE.2 2. 123, 135, 147, 159, 171, …

F.IF.3 3. f(1) = 123, f(n) = f(n – 1) + 12 for n > 1

F.LE.2 4. 128, 64, 32, 16, 8, …

F.IF.3 5.

F.LE.2 6. 8, –16, 32, –64, 128, …

F.LE.2 7. 72, 69, 66, 63, 60, ….

F.LE.2 8. D 96

F.LE.2 9. C

1 2 3 4 5

Term

Valu

e

Challenge Problem

F.IF.3 10. a. f(1) = 1, f(2) = 1, f(n) = 3[f(n – 2) + f(n – 1)] for n > 3

b. 81; 306; 1,161

73

− 79

f f n f n n( ) , ( ) ( ) •1 128 112

1= = − >for



ANSWERSLESSON 11: PUTTING IT TOGETHER

ANSWERS

F.IF.4 F.LE.2 F.LE.1

2.Type of Function General Form of the

EquationGeneral Form of the

Graph

Linear f(x) = mx + bx

y

Exponential f(x) = abxx

y

Quadratic f(x) = ax2 + bx + cx

y

(continues)



ANSWERSLESSON 11: PUTTING IT TOGETHER

(continued)

F.IF.4 F.LE.2 F.LE.1

2.Type of Function General Form of the

EquationGeneral Form of the

Graph

Arithmetic Sequence

Recursive Form: an = an–1 + d

where d is the constant difference and a is the term

x

y

Geometric Sequence

Recursive Form: Bn = cr(n–1)

where r is the constant ratio and c is the initial value

x

y

MODELING WITH FUNCTIONSrusdmath.weebly.com/uploads/1/1/1/5/11156667/g9_u1...High School: Modeling...

Documents

Transcript of MODELING WITH FUNCTIONSrusdmath.weebly.com/uploads/1/1/1/5/11156667/g9_u1...High School: Modeling...