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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
Copyright © 2015 Pearson Education, Inc. 41
High School: Modeling with Functions
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)
Copyright © 2015 Pearson Education, Inc. 42
High School: Modeling with Functions
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
Copyright © 2015 Pearson Education, Inc. 43
High School: Modeling with Functions
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
Copyright © 2015 Pearson Education, Inc. 44
High School: Modeling with Functions
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).
Copyright © 2015 Pearson Education, Inc. 45
High School: Modeling with Functions
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.
Copyright © 2015 Pearson Education, Inc. 46
High School: Modeling with Functions
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)
Copyright © 2015 Pearson Education, Inc. 47
High School: Modeling with Functions
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
Copyright © 2015 Pearson Education, Inc. 48
High School: Modeling with Functions
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.
Copyright © 2015 Pearson Education, Inc. 49
High School: Modeling with Functions
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
Copyright © 2015 Pearson Education, Inc. 50
High School: Modeling with Functions
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.
Copyright © 2015 Pearson Education, Inc. 51
High School: Modeling with Functions
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
Copyright © 2015 Pearson Education, Inc. 52
High School: Modeling with Functions
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
Copyright © 2015 Pearson Education, Inc. 53
High School: Modeling with Functions
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
F.IF.2 F.IF.3 F.LE.2
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
F.IF.2 F.IF.3 F.LE.2
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
Copyright © 2015 Pearson Education, Inc. 54
High School: Modeling with Functions
ANSWERSLESSON 9: RECURSIVE RULES FOR SEQUENCES
F.IF.2 F.IF.3 F.LE.2
9.
F.IF.3 F.LE.2
10. 1, 1, 2, 4, 7, 13, 24, 44, 81, 149
Challenge Problem
F.IF.2 F.IF.3 F.LE.2
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
Copyright © 2015 Pearson Education, Inc. 55
High School: Modeling with Functions
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
Copyright © 2015 Pearson Education, Inc. 56
High School: Modeling with Functions
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)
Copyright © 2015 Pearson Education, Inc. 57
High School: Modeling with Functions
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