TREY COX
SCOTT ADAMSON
COLLEGE ALGEBRA ACTIVITY GUIDE
TO ACCOMPANY COLLEGE ALGEBRA: A MAKE IT REAL APPROACH
College Algebra Activity Guide
For
College Algebra: A Make It Real Approach
Trey Cox
Chandler-Gilbert Community College
Scott Adamson
Chandler-Gilbert Community College
College Algebra Activity Table of Contents
Chapter 1 Activities
Activity 1.1 #1: Mathematical Modeling ...................................................................................... 1
Activity 1.1 #2: Mathematical Modeling ...................................................................................... 3
Activity 1.2 #1: Functions & Function Notation .......................................................................... 5
Activity 1.2 #2: Functions & Function Notation .......................................................................... 7
Activity 1.3 #1: Functions Represented by Tables ....................................................................... 9
Activity 1.3 #2: Functions Represented by Tables ......................................................................11
Activity 1.4 #1: Functions Represented by Graphs .....................................................................13
Activity 1.4 #2: Functions Represented by Graphs .....................................................................15
Activity 1.5 #1: Functions Represented by Words ......................................................................17
Activity 1.5 #2: Functions Represented by Words ......................................................................19
Activity 1.6 #1: Preview of Inverse Functions ............................................................................21
Activity 1.6 #2: Preview of Inverse Functions ............................................................................23
Chapter 2 Activities
Activity 2.1 #1: Functions with Constant Rates of Change .........................................................25
Activity 2.1 #2: Functions with Constant Rates of Change .........................................................27
Activity 2.2 #1: Modeling with Linear Functions .......................................................................29
Activity 2.2 #2: Modeling with Linear Functions .......................................................................31
Activity 2.3 #1: Linear Regression ..............................................................................................33
Activity 2.3 #2: Linear Regression ..............................................................................................35
Activity 2.4 #1: Systems of Linear Equations .............................................................................37
Activity 2.4 #2: Systems of Linear Equations .............................................................................39
Activity 2.5 #1: Systems of Linear Inequalities...........................................................................41
Activity 2.5 #2: Systems of Linear Inequalities...........................................................................43
Chapter 3 Activities
Activity 3.1 #1: Horizontal and Vertical Shifts ...........................................................................45
Activity 3.1 #2: Horizontal and Vertical Shifts ...........................................................................47
Activity 3.2 #1: Horizontal and Vertical Reflections ..................................................................49
Activity 3.2 #2: Horizontal and Vertical Reflections ..................................................................51
Activity 3.3 #1: Vertical Stretches and Compressions ................................................................53
Activity 3.3 #2: Vertical Stretches and Compressions ................................................................55
Activity 3.4 #1: Horizontal Stretches and Compressions ............................................................57
Activity 3.4 #2: Horizontal Stretches and Compressions ............................................................59
Chapter 4 Activities
Activity 4.1 #1: Variable Rates of Change ..................................................................................61
Activity 4.1 #2: Variable Rates of Change ..................................................................................63
Activity 4.2 #1: Modeling with Quadratic Functions ..................................................................65
Activity 4.2 #2: Modeling with Quadratic Functions ..................................................................67
Activity 4.3 #1: Quadratic Function Graphs and Forms ..............................................................69
Activity 4.3 #2: Quadratic Function Graphs and Forms ..............................................................71
Chapter 5 Activities
Activity 5.1 #1: Higher-Order Polynomial Function Modeling ..................................................73
Activity 5.1 #2: Higher-Order Polynomial Function Modeling ..................................................75
Activity 5.2 #1: Power Functions ................................................................................................77
Activity 5.2 #2: Power Functions ................................................................................................79
Activity 5.3 #1: Rational Functions .............................................................................................81
Activity 5.3 #2: Rational Functions .............................................................................................83
Chapter 6 Activities
Activity 6.1 #1: Percentage Change.............................................................................................85
Activity 6.1 #2: Percentage Change.............................................................................................87
Activity 6.2 #1: Exponential Function Modeling and Graphs .....................................................89
Activity 6.2 #2: Exponential Function Modeling and Graphs .....................................................91
Activity 6.3 #1: Compound Interest and Continuous Growth .....................................................93
Activity 6.3 #2: Compound Interest and Continuous Growth .....................................................95
Activity 6.4 #1: Solving Exponential and Logarithmic Equations ............................................ .97
Activity 6.4 #2: Solving Exponential and Logarithmic Equations ............................................ .99
Activity 6.5 #1: Logarithmic Function Modeling ...................................................................... 101
Activity 6.5 #2: Logarithmic Function Modeling ...................................................................... 103
Chapter 7 Activities
Activity 7.1 #1: Combinations of Functions .............................................................................. 105
Activity 7.1 #2: Combinations of Functions .............................................................................. 107
Activity 7.2 #1: Piecewise Functions......................................................................................... 109
Activity 7.2 #2: Piecewise Functions......................................................................................... 111
Activity 7.3 #1: Compositions of Functions .............................................................................. 113
Activity 7.3 #2: Compositions of Functions .............................................................................. 115
Activity 7.4 #1: Logistic Functions ........................................................................................... 117
Activity 7.4 #2: Logistic Functions ........................................................................................... 119
Activity 7.5 #1: Choosing a Mathematical Model ..................................................................... 121
Activity 7.5 #2: Choosing a Mathematical Model ..................................................................... 123
Chapter 8 Activities
Activity 8.1 #1: Using Matrices to Solve Linear Systems ......................................................... 125
Activity 8.1 #2: Using Matrices to Solve Linear Systems ......................................................... 127
Activity 8.2 #1: Matrix Operations and Applications ................................................................ 129
Activity 8.2 #2: Matrix Operations and Applications ................................................................ 131
Activity 8.3 #1: Using Inverse Matrices to Solve Matrix Equations ......................................... 133
Activity 8.3 #2: Using Inverse Matrices to Solve Matrix Equations ......................................... 135
Activity 1.1 #1: Mathematical Modeling
Activity Design:
Individual or Group
Activity Overview:
Create and analyze mathematical models in graphical form
Advance Preparation: None
1. Assuming that the buying power of $1.00 in the United States was actually $1.00 in 1982, the graph
shows the buying power for particular years. Year 0 corresponds to 1990. Describe the trend seen in
the graph of the buying power for the years shown.
The buying power of the dollar tends to decrease over the time period given. More specifically,
we see a fairly steady decrease from 1990 to 1996. There is a slight increase from 1997 to 1998
followed by another steady decrease from 1999 to 2001. A small increase is seen from 2001 to
2002 followed by another steady decrease in the buying power.
2. List each pair of consecutive years in which the buying power increased from one year to the next.
We see an increase in the buying power from 1997 to 1998 and again from 2001 to 2002.
3. Use the graph to estimate the amount of decrease in the power of the dollar from 1990 to 2004.
In 1990, the buying power is about $0.84. In 2004, the buying power is about $0.67. The
difference is 0.84 − 0.67 = 0.17 . The buying power of the dollar decreased by approximately
$0.17 from 1990 to 2004.
1
4. In which two consecutive years did the power of the dollar change the least? How can you tell?
Based on the graph, it appears that the buying power of the dollar changed the least from 1993
to 1994 and from 1996 to 1997. Between those years, the buying power appears to have
changed by $0.01 or less. The vertical distance between the points in those years appears to be
the smallest from among all the pairs of consecutive points.
5. Using the graph above to estimate, complete the table.
Year
Buying Power
(dollars)
1990
1995
1999
2001
2003
6. Referring to the table in (5), how many years does it appear to take for the value of the dollar to
decrease by $0.14?
In 1990, the buying power was $0.84 and in 2003 the buying power was $0.70. It took 13 years for
the buying power to drop by $0.14.
7. Referring to your observation in (6), predict when the buying power of the dollar will be $0? Does
this seem reasonable?
Yes. The buying power drops by $0.14 every 13 years. We can use a table to predict the buying
power of the dollar in the future.
Year
Buying Power
(dollars)
2003
2016
2029
2042
2055
2068
The buying power is predicted to by $0 in 2068. Even though the buying power will be less than
previous years, $0 doesn’t seem like a realistic value.
2
Activity 1.1 #2: Mathematical Modeling
Activity Design:
Individual or Group
Activity Overview:
Create and analyze mathematical models in tabular form
Advance Preparation: None
1. The data in the table show the number of golf courses in the United States for particular years
(Source: Statistical Abstract of the United States, 2004 – 2005, Table 1240). Calculate the year to
year change in the number of golf facilities. Is the number of golf facilities increasing or decreasing
from year to year? Explain.
t
Years
(Since 1980)
G
Golf Facilities
0
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
12005
12346
12384
12407
12582
12658
12846
13004
13210
13439
13683
14074
14341
14602
14900
15195
15489
15689
15827
15899
Difference
Since the differences are all positive, the number of golf courses is increasing from year to year.
However, the increase each year is not always the same.
2. Identify groups or individuals who could benefit from the analysis in (1).
This information could be important to golfers, the golf equipment industry, golf course
management schools, and advertisers because the increase in golf facilities corresponds to an
increase in people interested in golf.
3
3. The population of the United States and the number of golf facilities are given in the table. Compute
the number of people per golf facility for each of the given years. This value is found by dividing the
population by the number of golf facilities for the given year.
Year
1980
1990
1995
2000
2001
2002
2003
Population of the
United States
227,726,000
250,132,000
266,557,000
282,402,000
285,329,000
288,172,000
291,028,000
Number of Golf
Facilities
12005
12846
14074
15489
15689
15827
15899
Number of People
per Golf Facility
4. Is the increase in the number of golf facilities due to the increase in the population of the United
States? Use your computations from (3) to answer the question.
While the number of people per golf facility does fluctuate, the fluctuation is relatively small
considering the large population. We conclude that the increase in the number of golf facilities
has kept pace with the population of the United States since the number of people per golf
facility has stayed relatively constant.
5. Create a graph of the “Number of People per Golf Facility” (vertical axis) relative to the “Year”
(horizontal axis). Explain how the graph visually supports your answer to (4).
Between 1980 and 2003, the number of people per golf facility has ranged between roughly
18200 and 19500. The number of people per golf facility varied by about 1300 over the 23 year
time period.
4
Activity 1.2 #1: Functions & Function Notation
Individual or Group
Write and interpret functions using function notation and to evaluate and solve
function equations from real-world data
Advance Preparation: None
Activity Design:
Activity Overview:
The number of late airline arrivals (in thousands) in the United States from 1995 – 2005 and the
number of late departures (in thousands) is displayed in the table below. (Source: Statistical Abstract of
the United States 2007, Table 1054).
Late Airline Departures and Arrivals
from 1995 – 2005
d
a
Late Airline
Departures
(in 1000s)
827.9
846.9
870.4
937.3
1131.7
953.8
717.4
834.4
1187.6
1279.4
Late Airline Arrivals
(in 1000s)
1039.3
1083.8
1070.1
1152.7
1356.0
1104.4
868.2
1057.8
1421.4
1466.1
1. How many late airline arrivals were there when there were 1131.7 (thousand) late airline departures?
There were 1356.0 thousand late airline arrivals when there were 1131.7 thousand late airline
departures.
2. How many late airline departures were there when there were 1057.8 (thousand) late airline arrivals?
There were 834.4 thousand late airline departures when there were 1057.8 thousand late
airline arrivals.
3. What does the notation a = f (d ) mean in terms of airline departures and arrivals?
The notation a = f (d ) means that the number of late airline arrivals is a function of the
number of late airline departures.
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4. Solve f (d ) = 1466.1 for d and explain what the numerical value means in this context.
d is 1279.4 thousand late airline departures when there were 1466.1 thousand late airline arrivals.
5. Evaluate f (953.8) and explain what the numerical value means in this context.
The value of f (953.8) is 1104.4 and this means that when there are 953.8 thousand late airline
departures there were 1104.4 thousand late airline arrivals.
6. Estimate f (1300) and discuss the accuracy of your prediction.
An estimated value of f (1300) is 1500 (thousand). This estimate is not necessarily accurate
because we are extrapolating outside the given data so we don’t know the pattern outside the
given data. The estimate of 1500 means that when there are 1300 thousand late airline departures an
estimated 1500 thousand late airline arrivals occurred.
Based on the data from 1995 – 2005, the number of late airline arrivals in the United States (in
thousands) may be modeled by=
a (d ) 1.037 d + 167.9 late arrivals where d is the number of late
departures (in thousands).
7. Using the model=
a (d ) 1.037 d + 167.9 , evaluate a (900) and explain what the numerical value
means in this context.
We evaluate. a (900)
= 1.037(900) + 167.9
= 1101.2 . We estimate that there were 1101.2
thousand late airline arrivals when there were 900 thousand late airline departures.
8. Solve a (d ) = 900 for d and explain what the numerical answer represents in its real world context.
Write the solution in function notation.
To solve a (d ) = 900 for d we input 900 for the number of late airline arrivals and solve for d .
900 = 1.037d +167.9
900 - 167.9 = 1.037d
732.1= 1.037d
732.1
=d
1.037
705.98 = d
This means that when there were 900 thousand late airline arrivals there were 705.98 thousand
late airline departures and can be written in function notation as a (705.98) = 900 .
6
Activity 1.2 #2: Functions & Function Notation
Activity Design:
Individual or Group
Activity Overview:
Write and interpret functions using function notation and to evaluate and solve
function equations from real-world data
Advance Preparation: None
The number of juvenile aggravated assault arrests (in thousands) in the United States from 1999 – 2004
and the number of juvenile robbery arrests (in thousands) is displayed in the table below. (Source:
Statistical Abstract of the United States 2007, Table 318.)
Juvenile Arrests
from 1999 – 2004
r
a
Robbery
Arrests
(thousands)
Aggravated Assault
Arrests (thousands)
26,125
24,206
23,408
19,491
18,950
19,000
50,915
49,815
50,462
47,001
45,955
45,232
1. How many robbery arrests were there when there were 45,955 (thousand) aggravated assault arrests?
There were 18,950 (thousand) robbery arrests when there were 45,955 (thousand) aggravated
assault arrests.
2. How many aggravated assault arrests were there when there were 24,206 robbery arrests?
There were 49,815 (thousand) aggravated assault arrests when there were 24,206 (thousand)
robbery arrests.
3. What does the notation a  f (r ) mean in terms of aggravated assault arrests and robbery arrests?
The notation a  f (r ) means that the number of aggravated assault arrests is a function of the
number of robbery arrests.
4. Solve f (r )  50, 462 for r and explain what the numerical value means in this context.
r is 23,408 (thousand) robbery arrests when there were 50,462 (thousand) aggravated assault
arrests.
7
5. Evaluate f (24, 206) and explain what the numerical value means in this context.
The value of f (24, 206) is 49,815 and this means that when there were 24,206 (thousand) robbery
arrests there were 49,815 (thousand) aggravated assault arrests.
6. Estimate f (18,000) and discuss the accuracy of your prediction.
An estimate for f (18,000) is 42,000 (thousand) aggravated assault arrests for when there are
18,000 (thousand) robbery arrests. It is a challenge to arrive at an estimate for a value that is a
so far away from that given and may not be likely to be accurate.
Based on the data from 1999 – 2004, the number of juvenile aggravated assault arrests (in thousands)
in the United States may be modeled by a(r )  0.7960r  30,830 where r is the number of juvenile
robbery arrests (in thousands).
7. Using the model a(r )  0.7960r  30,830 , evaluate a(30,000) and explain what the numerical value
means in this context.
We evaluate. a(30,000)  0.7960(30,000)  30,830  54,710 . We estimate that there were
54,710 (thousand) aggravated assault arrests when there were 30,000 (thousand) juvenile
robbery arrests.
8. Solve a(r )  46,000 for r and explain what the numerical answer represents in its real world
context. Write the solution in function notation.
To solve a(r )  46,000 for r we input 46,000 for the number of aggravated assault arrests
and solve for r .
46, 000 = 0.7960r  30,830
46, 000  30,830 = 0.7960r
15170 = 0.7960r
15170
=r
0.7960
19058 = r
This means that when there are 46,000 (thousand) aggravated assault arrests there will be
19,057 (thousand) juvenile robbery arrests and can be written in function notation as
a(19,057)  46,000 .
8
Activity 1.3 #1: Functions Represented by Tables
Individual or Group
Calculate average rates of change from tables and use to estimate unknown
function values
Advance Preparation: None
Activity Design:
Activity Overview:
The number of daylight hours on the 21st day of each month in Anchorage, Alaska is displayed in the
table below. (Source: www.aa.usno.navy.mil.)
Daylight Hours on the 21st day
of the month in Anchorage, Alaska
m
h
Month
Daylight Hours
1
3
5
7
9
11
6.87
12.33
17.98
18.00
12.35
6.78
( m = 1 is January)
1. Calculate the average rates of change for consecutive pairs of data values.
Daylight Hours on the 21st day
of the month in Anchorage, Alaska
m
Month
h
( m = 1 is January)
Daylight Hours
1
6.87
3
12.33
5
17.98
7
18.00
9
12.35
11
6.78
Average Rate of Change
2. Explain the meaning of the rates of change in the context of the data.
For each two month interval the rate of change tells how many more daylight hours (if the rate is
positive) or fewer (if the rate is negative) there are per month on average.
9
3. Estimate h(4) and interpret what it means in the real world context.
Since the number of daylight hours in Anchorage for the third month is 12.33 we can add the
average rate of change of approximately 2.83 hours to get an estimate of 15.16 hours for the
number of daylight hours there will be on April 21st ( m = 4 ).
4. Estimate the number of daylight hours on December 21 and write your answer in function notation.
Since December 21 falls between November 21 and January 21 of the following year, we need to
calculate the average rate of change between those two months.
6.87 − 6.78
= 0.045
2
Since the number of daylight hours in Anchorage for the eleventh month is 6.78 we can add the
average rate of change of approximately 0.05 hours get an estimate of 6.83 hours for the number of
daylight hours there will be on December 21 ( m = 12 ).
5. December 21 is the shortest day of the year and June 21 is the longest day of the year in the northern
hemisphere. Determine the months when the number of daylight hours is increasing. Support your
decision with average rates of change and by relating your answer to what you would expect in the
real world.
It appears from the positive rates of change for months 1 through 6 that the number of daylight
hours would be increasing from January through June. This would make sense because Anchorage
is in the northern hemisphere and in the spring time daylight hours would increase.
6. Determine the months when the number of daylight hours is decreasing. Support your decision with
average rates of change and by relating your answer to what you would expect in the real world.
It appears from the negative rates of change for months 6 through 12 that the number of daylight
hours would be decreasing from July through December. This would make sense because
Anchorage is in the northern hemisphere and in the winter time daylight hours would decrease.
7. Calculate the average rate of change between months m = 1 and m = 5 and explain what the
numerical value means in this context and whether or not your answer seems reasonable.
17.98 − 6.87
= 2.78
5 −1
This tells us that on average 2.78 more hours of daylight are gained each month from month 1
to month 5. This makes sense because this is during the Spring and Summer so the length of
day continue to increase over that time span in the Northern Hemisphere.
8. Calculate the average rate of change between months m = 3 and m = 9 and explain what the
numerical value means in this context and whether or not your answer seems reasonable.
12.35 − 12.33
= 0.003
9−3
This tells us that on average 0.003 more hours of daylight are gained each month from month 3
to month 9. This makes sense because the months 3 (March) and 9 (September) are in Spring
and Fall and the overall change between them would not be great. Even though the number of
daylight hours increase throughout the Spring and Summer they likewise decrease throughout
Fall and Winter to result in little or no overall change.
10
Activity 1.3 #2: Functions Represented by Tables
Individual or Group
Calculate average rates of change from tables and use to estimate unknown
function values
Advance Preparation: None
Activity Design:
Activity Overview:
The number of daylight hours on the 21st day of each month in Seattle, Washington is displayed in the
table below. (Source: www.aa.usno.navy.mil.)
Daylight Hours on the 21st day of each
month in Seattle, Washington
m
h
Month
(Jan is m = 1 )
Daylight Hours
1
3
5
7
9
11
9.10
12.22
15.38
15.40
12.23
9.05
1. Calculate the average rates of change for consecutive pairs of data values.
Daylight Hours on the 21st day of the month in Seattle,
Washington
m
h
Month (Jan is
Daylight Hours
1
9.10
3
12.22
5
15.38
7
15.40
9
12.23
11
9.05
m = 1)
Average Rate of Change
2. Explain the meaning of the rates of change in the context of the data.
For each two hour interval the rate of change tells how many more (if the rate is positive) or
fewer (if the rate is negative) daylight hours there are per month on average.
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3. Estimate h(4) and interpret what it means in the real world context.
Since the number of daylight hours in Seattle for the third month is 12.22 we can add the
average rate of change of 1.58 hours to find an estimate of 13.80 for how many daylight hours
there will be on April 21st ( m = 3 ).
4. Estimate the number of daylight hours on October 21st and write your answer in function notation.
Since the number of daylight hours in Seattle for the ninth month is 12.23 we can subtract the
average rate of change of 1.59 hours to find an estimate of 10.64 for how many daylight hours
there will be on October 21st ( m = 10 ). This can be written h(10) = 10.64 .
5. Determine the months when the number of daylight hours is increasing. Support your decision with
average rates of change and by relating your answer to what you would expect in the real world.
It appears from the positive rates of change for months 1 through 5 that the number of daylight
hours would be increasing from December through June. This would make sense because
Seattle is in the Northern Hemisphere and in the spring time daylight hours would increase.
6. Determine the months when the number of daylight hours is decreasing. Support your decision with
average rates of change and by relating your answer to what you would expect in the real world.
It appears from the negative rates of change for months 7 through 11 that the number of
daylight hours would be decreasing from June through December. This would make sense
because Seattle is in the Northern Hemisphere and in the winter time daylight hours would
decrease.
7. Calculate the average rate of change between months m = 1 and m = 5 and explain what the
numerical value means in this context and whether or not your answer seems reasonable.
15.38 − 9.10
= 1.57
5 −1
This tells us that on average 1.57 more hours of daylight are gained each month from month 1
to month 5. This makes sense because this is during the Spring and Summer so the length of
day continue to increase over that time span in the Northern Hemisphere.
8. Calculate the average rate of change between months m = 3 and m = 9 and explain what the
numerical value means in this context and whether or not your answer seems reasonable.
12.23 − 12.22
= 0.002
9−3
This tells us that on average 0.002 more hours of daylight are gained each month from month 3
to month 9. This makes sense because the months 3 (March) and 9 (September) are in Spring
and Fall and the overall change between them is not great. Even though the number of daylight
hours increase throughout the Spring and Summer they would likewise decrease throughout
Fall and Winter to result in little or no overall change.
12
Activity 1.4 #1: Functions Represented by Graphs
Activity Design:
Activity Overview:
Individual or Group
Find function values, estimate practical domain and practical range, and
interpret the real-world meaning of a graph and characteristics of a graph
(vertical, horizontal intercepts, slope)
Advance Preparation: None
The total number of sea otters, s , counted at Point Lobos, California in May of each year y since
1980 by the United States Geological Survey is displayed in the graph of the function s( y ) below.
(Source: Modeled from http://pt-lobos.parks.state.ca.us/2007_Fall.pdf .)
1. Determine the practical domain and practical range for s( y ) and explain your answer in terms of
this context.
According to the graph, the practical domain is 0  y  27 . This means that the model makes
sense for years between 1980 and 2007. Since the graph doesn’t show the sea otter population
beyond this interval of time, other input values have unknown outputs.
2. Using the notion of rate of change, explain what the graph tells us in its real-world context.
In 1980 the number of sea otters was about 1400 and decreasing (negative rate of change). The
population decreased until about 1983. The number of sea otters increased from 1983 until
about 1995 and then decreased a little until 2000 and then increased dramatically.
3. Estimate s(15) and interpret what it means in the real world context.
The number of sea otters in year 15 (1995) is about 2250.
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4. Use the graph to estimate the solution to s( y)  1400 and explain what your solution means in the
real-world context.
It appears that in both 1980 and 1985 the number of sea otters was 1400.
5. Estimate the vertical intercept of the function s( y ) and explain what the coordinates of this point
mean in its real-world context.
The coordinates of the vertical intercept are about (0,1400) which means that in 1980 the
number of sea otters was about 1400.
6. If there was a horizontal intercept for the function s( y ) explain what the coordinates of this point
would mean in its real-world context.
The coordinates of the horizontal intercept would be of the form  a, 0  . This would mean that
in year a there would be no sea otters at Point Lobos, California.
7. Estimate the coordinates of the point on the graph where you believe the slope of the function is
greatest explain what this point means in this context.
An estimate for the coordinates would be (27,3500) . This means that in 2007 the number of
sea otters was 3500 and was increasing at the fastest pace since 1980.
8. Estimate the coordinates of a point on the graph where the slope of the function is negative and
explain what this point means in this context.
An estimate for the coordinates a point is (1,1300) . This means that in 1981 the number of sea
otters was 1300 and was decreasing.
14
Activity 1.4 #2: Functions Represented by Graphs
Activity Design:
Activity Overview:
Individual or Group
Find function values, estimate practical domain and practical range, and
interpret the real-world meaning of a graph and characteristics of a graph
(vertical, horizontal intercepts, slope)
Advance Preparation: None
Based on data from 1985  2003, the number of cases of Chicken Pox in the United States may be
modeled by the graph of C (t ) in thousand cases where t is the number of years since 1985. (Source:
Modeled from Statistical Abstract of the United States 2006, Table 177.)
1. Determine the practical domain and practical range for C  t  and explain your answer in terms of
this context.
The practical domain is likely to be from about 0 (1985) or a few years before to 20 (2005) or
only a few years after because we don’t have enough information about what happened before
1985 to extrapolate into the past. We can make the assumption that this graphical pattern
continues into the future but that would mean that the number of chicken pox cases would level
off and not rise or fall.
2. Using the notion of rate of change, explain what the graph tells us in its real-world context.
Starting in 1985 the number of chicken pox cases was about 180,000 and then the rate of
change became negative meaning that the number dropped significantly until about year 14
(1999) when the rate of change began to approach zero meaning that the number of chicken
pox cases began to level off.
3. Estimate C (15) and interpret what it means in the real world context.
The number of chicken pox cases in year 15 (2000) was about 30,000.
15
4. Use the graph to estimate the solution to C (t )  100 and explain what your solution means in the
real-world context.
It appears that in year 11 (1996) the number of chicken pox cases was 100,000.
5. Estimate the vertical intercept of the function C  t  and explain what the coordinates of this point
mean in its real-world context.
The coordinates of the vertical intercept are about (0,178) which means that in 1985 the
number of chicken pox cases was about 178,000.
6. If there was a horizontal intercept for the function C  t  explain what the coordinates of this point
would mean in its real-world context.
The coordinates of the horizontal intercept would have a C -coordinate of 0 and a first t coordinate that is not 0. This would tell us in which year there would be no chicken pox cases.
7. Estimate the coordinates of the point on the graph where you believe the slope of the function is
most negative explain what this point means in this context.
An estimate for the coordinates is (11,30) . This means that in year 11 (1996) the number of
chicken pox cases was 100,000 and was decreasing at the fastest pace during the interval of
time from 1985 to 2003.
8. Estimate the coordinates of a point on the graph where the slope of the function is near 0 (not
negative or positive) and explain what this point means in this context.
An estimate for the coordinates would be (0,180) or (18, 20) . This means that in 1985 the
number of chicken pox cases in the year 1985 was 180,000 and not changing much. Also in the
year 2003 the number of chicken pox cases was 20,000 and not changing much.
16
Activity 1.5 #1: Functions Represented by Words
Individual or Group
Convert graphical and tabular functional representations into a verbal
description
Advance Preparation: None
Activity Design:
Activity Overview:
The rough sketch below describes what happens in a hypothetical 400 meter race of runners A, B and
C.
1. Imagine that you are the race commentator. Describe (in writing) what is happening during the race
with Runner A.
Runner A starts out quickly from the starting block and continually slows down as the race
goes on.
2. Imagine that you are the race commentator. Describe (in writing) what is happening during the race
with Runner B.
Runner B starts out slowly from the starting block and continually speeds up as the race goes
on.
3. Imagine that you are the race commentator. Describe (in writing) what is happening during the race
with Runner C.
Runner C starts out faster than B but slower than A from the starting block and runs at a
constant speed throughout the race.
17
4. Estimate the vertical intercept for the three runners and explain what the coordinates of each point
means in its real-world context.
It appears that the vertical intercept for all three runners is (0, 0) which means at time 0
seconds the racers are all at 0 meters.
5. Estimate the horizontal intercept for the three runners and explain what the coordinates of each point
means in its real-world context.
It appears that the horizontal intercept for all three runners is (0, 0) which means at time 0
seconds the racers are all at 0 meters.
6. Explain what happens in the race when the graph for Runner B and Runner C intersect.
At that time (about 65 seconds) both runners B and C are the same distance from the starting
line. (Note: They are not colliding!)
7. Do all three runners finish the 400 meter race within 70 seconds? Explain.
No, all three runners do not complete the 400 meters within 70 seconds. It appears from the
graph that only Runner A does as the graph for Runner A has already reached 400 meters by
the 70 second mark while the others have not.
8. In what place (1st, 2nd, and 3rd) does each runner come in? Explain your reasoning by using the
graph.
Runner A earned 1st, Runner B earned 2nd, and Runner C earned 3rd. This is because Runner
A’s graph has a second coordinate of 400 at the earliest time, Runner B’s graph has a second
coordinate of 400 second, and Runner C’s graph has a second coordinate of 400 meters last.
18
Activity 1.5 #2: Functions Represented by Words
Individual or Group
Convert graphical and tabular functional representations into a verbal
description
Advance Preparation: None
Activity Design:
Activity Overview:
The graphs below display the years of life expectancy depending upon what year a person is born.
(Source: Modeled from Statistical Abstract of the United States, 2006, Table 96)
1. Describe (in writing) what the graph tells us about life expectancy of women.
Women’s life expectancy is increasing at a relatively constant rate.
2. Describe (in writing) what the graph tells us about life expectancy of men.
Men’s life expectancy is increasing at a relatively constant rate.
3. Estimate the vertical intercept for women and explain what the coordinates of the point means in its
real-world context.
It appears that the vertical intercept for women is about (0, 77.8) which means in year 0 or
1980 women’s life expectancy was 77.8 years.
19
4. Estimate the vertical intercept for men and explain what the coordinates of the point means in its
real-world context.
It appears that the vertical intercept for men is about (0, 70.1) which means in year 0 or 1980
men’s life expectancy was 70.1 years.
5. If there were a horizontal intercept for men or women explain what the coordinates of each point
would mean in this real-world context.
If there were a horizontal intercept that would mean that in a particular year the life
expectancy would be 0 years. In other words as soon as someone was born they would die!
6. If the trends/patterns displayed in the graphs were to continue, does it appear that the graphs would
intersect? Explain what the point of intersection would mean in this context.
Yes. The graph of the men’s life expectancy is steeper telling us that their life expectancy is
increasing at a faster rate and therefore would catch that of women’s at some point in the
future if the trend continued long enough.
7. Do women ever have a life expectancy of 70 years? Explain.
Not in the interval of years given in the graph but if we went into the past and continued the
trend then women would have had a life expectancy of 70 years at some point before 1980.
8. The graph for men’s is steeper than that of the women’s. What does this mean?
This tells us that men’s life expectancy is increasing at a faster rate than women’s.
20
Activity 1.6 #1: Preview of Inverse Functions
Activity Design:
Individual or Group
Activity Overview:
Describe and use inverse function notation
Advance Preparation: None
Based on data from 2003, the percent of Americans using a computer at work may be modeled by
P  f ( I )  0.5530I  29.62 where I is the family income in thousand dollars. (Source: Modeled
from National Center for Education Statistics, Digest of Education Statistics: 2005, Table 421).
1. Find P  f (50) and explain what it means.
P  f (50)  0.553050  29.62  57.27 percent. This means that about 57.27% of Americans
with a family income of $50,000 use a computer at work.
2. Interpret the meaning of I  f 1 ( P) . No calculations are needed.
I  f 1 ( P) takes the percentage of Americans using a computer at work as the input and
outputs the family income in thousands of dollars. This is the inverse function since the original
function, P  f (I ) , takes the income in thousands of dollars as the input and outputs the
percentage of Americans using a computer at work.
3. Complete the table.
Income as a Function of
Percentage of Computer Use
I  f 1 ( P)
I
P
Income
Percentage of
(thousands of
Computer Use
dollars)
Percentage of Computer Use as
a Function of Income
P  f (I )
I
P
Income
Percentage of
(thousands of
Computer Use
dollars)
10
35.15
20
40.68
30
46.21
40
51.74
50
57.27
60
62.80
70
68.33
80
73.86
90
79.39
21
4. Use the table from #3 to compute I  f 1 (40.68) then explain what it means.
I  f 1 (40.68)  20 (in thousands of dollars). This tells us that 40.68% of Americans with a
family income of $20,000 use a computer at work.
Use the graph of P  f (I ) to respond to the following items.
5. Estimate P  f (30) and explain its meaning. Explain how to read the graph to make this estimation.
P  f (30)  46% . This means that about 46% of Americans with a family income of $30,000
use a computer at work. We find this result by first locating 30 (in thousands of dollars) on the
horizontal axis. Moving vertically until we reach the graph of the function, we then read across
horizontally to the vertical scale and see the corresponding percentage of about 46%.
6. Estimate I  f 1 (80) and explain its meaning. Explain how to read the graph to make this
estimation.
I  f 1 (80)  90 (in thousands of dollars). About 80% of Americans with a family income of
$90,000 use a computer at work. We find this result by first locating 80 percent on the vertical
axis. Moving horizontally until we reach the graph of the function, we then read down
vertically to the horizontal scale and see the corresponding income of about $90,000.
22
Activity 1.6 #2: Preview of Inverse Functions
Activity Design:
Individual or Group
Activity Overview:
Describe and use inverse function notation
Advance Preparation: None
The graph shows the relationship between the price of a DVD player and the number of DVD units
sold (Source: Consumer Electronics Association-www.ce.org). We will model this situation using the
function P = f (U ) where P is the DVD price in dollars and U is the number of DVD units sold, in
thousands, at price P .
1. Estimate P = f (5000) and explain what it means.
P = f (5000) ≈ 230 (in dollars). This means that at a price of $230, 5000 (thousand) DVD
units are sold.
2. Interpret the meaning of U = f
−1
( P) . No calculations are needed.
The inverse function U = f −1 ( P) takes the DVD player price as the input and outputs the
number of DVD units sold (in thousands) at that price. This is the inverse of P = f (U ) where
the number of DVD units sold (in thousands) is the input and the output is the price of the DVD
player.
23
3. Complete the table by estimating values from the graph.
Number of DVD Units Sold as
a Function of DVD Player
Price
U = f −1 ( P )
U
P
DVD Units
DVD Player
Sold
Price (est.)
(thousands)
DVD Player Price as a
Function of Number of DVD
Units Sold
P = f (U )
U
P
DVD Units
DVD Player
Sold
Price (est.)
(thousands)
0
285
2500
255
5000
230
7500
210
10000
190
12500
175
15000
155
17500
140
20000
130
4. Use the table from #3 to compute U = f −1 (155) then explain what it means.
U = f −1 (155) ≈ 15000 . This tells us that at a price of $150, 15000 (thousand) DVD units may
be sold.
5. Use the table from #3 to create a graph of U = f
−1
24
( P) .
Activity 2.1 #1: Functions with Constant Rates of Change
Activity Design:
Individual or Group
Activity Overview:
Explain the practical meaning of slope, horizontal intercept, vertical intercept
of functions with a constant rate of change.
Advance Preparation: Stopwatch
Set-Up:
Start with two people standing face-to-face holding hands (like at a wedding). A third student can be
the time keeper. When the time keeper says “go,” the designated squeeze starter squeezes one of the
hands of the second person. Upon feeling the squeeze, the second person squeezes back with the other
hand. The squeeze starter shouts “Stop!” when the squeeze is felt and the time keeper stops the
stopwatch. Record the time. Continue this procedure by adding one person to the circle each iteration.
The data in the table below is generated from the model created to represent the reaction times from a
class activity performed by the author in Spring 2004.
Sample Squeeze Data
p
People in
Circle
T
Time (seconds)
2
4
6
8
10
1.08
1.56
2.04
2.52
3.00
1. Find the rate of change between consecutive data points in the table.
Sample Squeeze Data
p
People in
Circle
T
Time (seconds)
2
1.08
4
1.56
6
2.04
8
2.52
10
3.00
25
Rate of Change
2. Provide a detailed interpretation of the meaning of the rate of change.
The rate of change for each consecutive data point is 0.24 which tells us that for every increase
of 1 person in the circle the reaction time is increased by 0.24 seconds.
3. Estimate the reaction time for a circle of 7 people.
The reaction time is 2.04 seconds for a circle of 6 people so we must add 0.24 seconds in order
to estimate the reaction time when the circle has 6. Therefore the answer is 2.28 seconds.
4. Estimate the reaction time for a circle of 14 people.
The reaction time is 3.00 seconds for a circle of 10 people so we must add 0.24 seconds four
times in order to estimate the reaction time when the circle has 14. Therefore the answer is
3.00 + 0.24 + 0.24 + 0.24 + 0.24 = 3.96 seconds.
5. When the reaction time is 5 seconds how many people would you estimate to be in the circle?
The reaction time is 3.00 seconds for a circle of 10 people so we must add 0.24 seconds enough
times in order to reach the reaction time of 5 seconds.
3.00  0.24  0.24  0.24  0.24  0.24  0.24  0.24  0.24  4.92
We will need to add approximately eight more people so there will be a total of 18 people in the
circle for a reaction time of 5 seconds.
6. Create a function, T ( p) , to model the reaction time as a function of the people in the circle.
We know that the formula y  b  mx models functions with a constant rate of change. We
know that the reaction time is 1.08 seconds for a circle of 2 people. We don’t know what the
reaction time would be for a hypothetical circle of size 0 but we can figure out what it would be
by subtracting 0.24 repeatedly. We get 1.032 seconds. The rate of change is 0.24 seconds per
person. This gives us the linear function T ( p)  1.032  0.24 p .
7. Is the slope of the line, T ( p) , positive, negative, or zero. Explain.
The slope is positive because the rate of change is positive telling us that the reaction time will
increase for every addition person in the circle.
8. Graph the function T ( p) and label the axes with appropriate values.
26
Activity 2.1 #2: Functions with Constant Rates of Change
Activity Design:
Individual or Group
Activity Overview:
Explain the practical meaning of slope, horizontal intercept, vertical intercept
of functions with a constant rate of change.
Advance Preparation: None
The U.S. Government’s Special Supplemental Nutrition Program for Women, Infants, and Children, is
commonly referred to as WIC. WIC gives grants to states for supplemental foods, nutrition education,
and medical referrals for low-income mothers and their young children. To qualify for the program, the
household income must fall below certain limits as shown in the table below effective from July 1st,
2007 to June 30th, 2008. (Source: www.fns.usda.gov )
48 Contiguous States
m
A
Members of
Annual Income (dollars)
Household
1
2
3
4
5
18,889
25,327
31,765
38,203
44,641
1. Find the rate of change between consecutive data points in the table.
48 Contiguous States
m
A
Members of
Annual Income (dollars)
Household
1
18,889
2
25,327
3
31,765
4
38,203
5
44,641
27
Rate of Change
2. Provide a detailed interpretation of the meaning of the rate of change.
The rate of change for each consecutive data point is 6438 which tells us that for every
increase of 1 household member the annual income allowed is increased by $6438.
3. Estimate the allowed annual income for a household of 6.
The allowable income is $44,641 for a household of 5 so we must add $6438 in order to
estimate the annual salary when the household is 6. Therefore the answer is $51,079.
4. Estimate the allowed annual income for a household of 8.
The allowable income is $44,641 for a household of 5 so we must add $6438 three times in
order to estimate the annual salary when the household is 8. Therefore the answer is
$44,641 + $6438 + $6438 + $6438 = $63,955.
5. When the annual income is $40,000 how many household members can there be to qualify?
The maximum annual income of $38,203 is allowed for a household of 4 but for a household of
5 the annual income can reach $44,641 so there would have to be a household size of 5 for a
$40,000 annual income.
6. Create a function, A(m) , to model the annual income as a function of the members of the
household.
We know that the formula y= b + mx models functions with a constant rate of change. We
know that the annual income is $18,889 for a household of 1. We don’t know what the annual
income is for a hypothetical household of size 0 but we can figure out what it would be by
subtracting $6438. We get $12,541. The rate of change is $6438. This gives us the linear
function =
A(m) 12, 451 + 6438m .
7. Is the slope of the line, A(m) , positive, negative, or zero. Explain.
The slope is positive because the rate of change is positive telling us that the annual salary will
increase for every household member increase.
8. Graph the function A(m) and label the axes with appropriate values.
28
Activity 2.2 #1: Modeling with Linear Functions
Activity Design:
Individual or Group
Activity Overview:
Determine the slope, intercepts, and make predictions from a linear function.
Advance Preparation: None
Medical researchers have found that there is a linear relationship between a person’s blood pressure
and their weight. In males 35 years of age, for every 5 pound (lb.) increase in a person’s weight there is
generally an increase in the systolic blood pressure of 2 millimeters of mercury (mmHg). Moreover,
for a male 35 years of age and 190 pounds the preferred systolic blood pressure is 125. (Source:
www.ash-us.org ).
1. Construct a formula for a 35-year-old male’s systolic blood pressure as a function of his weight,
B ( w) .
2 mmHg
m=
5 lb
2
B(=
w)
w+b
5
2
125
(190) + b
=
5
49 = b
2
B(=
w)
w + 49
5
2. Give a practical domain and range for the function B( w) .
The practical domain could be (100, 260) because the weights of most 35-year-old males are in
that interval. Based upon the practical domain, the practical range would be (89, 129).
3. Explain the meaning of the slope in the real world context.
2 mmHg
5 lb
For every 5 pound (lb.) increase in a person’s weight there is generally an increase in the
systolic blood pressure of 2 millimeters of mercury (mmHg)
m=
4. Determine the coordinates of the vertical intercept and explain its hypothetical meaning in the real
world context.
2
B=
(0)
(0) + 49
5
B (0) = 49
When the male weighs 0 pounds, his blood pressure is 49 mmHg.
29
5. Determine the coordinates of the horizontal intercept and explain its hypothetical meaning in the real
world context.
=
0 0.4 w + 49
−49 =
0.4 w
−49
=w
0.4
−122.5 =
w
The horizontal intercept means that the blood pressure is 0 mmHg if the man weighed 122.5 pounds.
6. Using the function B( w) , predict the blood pressure for a 200 pound 35-year-old male.
=
B (200) 0.4(200) + 49
B (200) = 129
The blood pressure is 129 mmHg for a 200 pound man.
7. Evaluate B(176) and explain its meaning in the real world context.
=
B (176) 0.4(176) + 49
B (176) = 119.4
The blood pressure is 119.4 mmHg for a 176 pound man.
8. Solve B ( w) = 114 and explain its meaning of in the real world context.
=
114 0.4( w) + 49
65 = 0.4 w
65
=w
0.4
162.5 = w
The weight is 162.5 pounds if the blood pressure is 114 mmHg.
9. Graph the function B( w) and label the axes with appropriate values.
30
Activity 2.2 #2: Modeling with Linear Functions
Activity Design:
Individual or Group
Activity Overview:
Determine the slope, intercepts, and make predictions from a linear function.
Advance Preparation: None
Based on data from 1970 − 2004, the production of crude oil may be modeled by
P (t ) =
−0.27t + 20.67 in quadrillion barrels where t is the number of years since 1970 (Source:
Modeled from Statistical Abstract of the United States, 2007, Table 895).
1. Describe what the function P (t ) tells us in terms of the real world context.
As the years increase, the oil production decreases at a constant rate.
2. Give the practical domain and range for the function P (t ) .
The practical domain could be from (−10, 40) because we need to be careful how far we make
predictions into the past and into the future since situations change. Based upon the practical
domain, the practical range would be from (9.87, 17.97).
3. Explain the meaning of the slope in the real world context.
The oil production drops 0.27 quadrillion barrels each year.
4. Determine the coordinates of the vertical intercept and explain its meaning in the real world context.
P(0) =
−.27(0) + 20.67
P(0) = 20.67
In year 0 or in 1970, the oil production would be 20.67 quadrillion barrels.
5. Determine the coordinates of the horizontal intercept and explain its meaning in the real world
context.
−.27t + 20.67
0=
−20.67 =
−.27t
−20.67
=t
−.27
76.66 = t
The model estimates that oil production is 0 quadrillion barrels in the year 76 or in 2047.
31
6. Using the function P (t ) , predict the oil production in the year 2009.
P(39) =
−0.27(39) + 20.67
P(39) = 10.14
In 2009, the oil production is estimated to be 10.14 quadrillion barrels.
7. Evaluate P(29) and explain its meaning of in the real world context.
P (29) =
−0.27(29) + 20.67
P (29) = 12.84
In 1999, the oil production is estimated to be 12.84 quadrillion barrels.
8. Solve P (t ) = 16 and explain its meaning of in the real world context.
We make the oil production to be 16 quadrillion barrels.
−0.27t + 20.67
16 =
−4.67 =
−0.27t
−4.67
=t
−0.27
17.30 = t
When 16 quadrillion barrels are produced, the year is 1987.
9. Graph the function P (t ) and label the axes with appropriate values.
32
Activity 2.3 #1: Linear Regression
Individual or Group
Use linear regression to make predictions and explain the strength of fit of the
regression model.
Advance Preparation: None
Activity Design:
Activity Overview:
The Washington State Higher Education Coordinating Board, Higher Education Statistics, reports the
following Washington State Public University Enrollment (Source: www.hecb.wa.gov).
Washington State Public University
Enrollment
S
t
Students
Years
(in thousands)
(Since 1990)
7
8
9
10
11
12
13
14
15
80.6
82.0
82.8
84.8
88.0
89.5
90.1
91.4
91.6
1. Suppose the linear model shown in the graph above is created for these data. Compute the total sum
of squares.
t
Years
(Since 1990)
7
8
9
10
11
12
13
14
15
S
Students
(in thousands)
80.6
82.0
82.8
84.8
88.0
89.5
90.1
91.4
91.6
Linear Model
(students in 1000’s)
y = 1.375 x + 70.975
Residual
Square of
Residual
Total Sum of
Squares of
Residuals
The total sum of squares for y = 1.375 x + 70.975 is approximately 11.
33
2. Use a calculator to determine a linear regression model for these data. Explain what the coefficient
of determination and the correlation coefficient tells you about the strength of fit of this model.
y = 1.525 x + 69.9806 , r 2 ≈ 0.96 , r ≈ 0.98 . The coefficient of determination is relatively close
to 1 and therefore tells us that the model fits the data well. The correlation coefficient is
relatively close to positive 1 also telling us that the model fits the data well. Since r > 0 , we
know that the function is increasing. That is, as time increases, so does the total enrollment.
3. Calculate the total sum of the squares of the residuals of the regression model to show that it is less
than that found in #1.
The total sum of squares is approximately 5.7 which is less than the total sum of squares seen
with the model presented in #1.
t
Years
(Since 1990)
7
8
9
10
11
12
13
14
15
S
Students
(in thousands)
80.6
82.0
82.8
84.8
88.0
89.5
90.1
91.4
91.6
Linear Model
(students in 1000’s)
y = 1.525 x + 69.9806
Residual
Square of
Residual
Total Sum of
Squares of
Residuals
4. Interpret the meaning of the slope and the vertical intercept of the model in terms of enrollment, E ,
and years since 1990, t .
The slope of 1.525 tells us that for each additional year, enrollment increases by 1.525
(thousand) students. The vertical intercept tells us that in 1990, the model predicts the
enrollment to be about 69.98 (thousand) students.
5. Use the model to predict public university enrollment in Washington in 2010. Is this an example of
interpolation or extrapolation? Why?
y = 1.525(20) + 69.9806 ≈ 100.48 thousand students. The regression model predicts that in
2010, 100,480 students will be enrolled in universities in Washington. This is an example of
extrapolation since the prediction is outside of or beyond the given data set.
34
Activity 2.3 #2: Linear Regression
Activity Design:
Activity Overview:
Individual or Group
Use linear regression to make predictions and explain the strength of fit of the
regression model.
Advance Preparation: None
The number of adults in prison, jail, on probation, or on parole is provided in the table (Source:
Statistical Abstract of the United States, 2008; Table 338).
Adults in Prison, in Jail, on
Probation, or on Parole
t
Years
(Since 1990)
A
Adults
(in millions)
0
2
4
6
8
10
12
14
4.4
4.8
5.1
5.5
6.1
6.4
6.8
7.0
1. Suppose the linear model shown in the graph above is created for these data. Compute the total sum
of squares.
t
Years
(Since 1990)
A
Adults
(in millions)
0
2
4
6
8
10
12
14
0
4.4
4.8
5.1
5.5
6.1
6.4
6.8
7.0
4.4
Linear Model
(Adults in millions)
y  0.186 x  4.4
Residual
Total Sum of
Squares of
Residuals
The total sum of squares for y  0.186 x  4.4 is approximately 0.096.
35
Square of
Residual
2. Use a calculator to determine a linear regression model for these data. Explain what the coefficient
of determination and the correlation coefficient tells you about the strength of fit of this model.
y  0.195x  4.4 , r 2  0.992 , r  0.996 . The coefficient of determination is relatively close to
1 and therefore tells us that the model fits the data well. The correlation coefficient is relatively
close to positive 1 also telling us that the model fits the data well. Since r  0 , we know that the
function is increasing. That is, as time increases, so does the total number of people in prison,
jail, on probation or on parole.
3. Calculate the total sum of the squares of the residuals of the regression model to show that it is less
than those found in #1.
t
Years
(Since 1990)
A
Adults
(in millions)
0
2
4
6
8
10
12
14
0
4.4
4.8
5.1
5.5
6.1
6.4
6.8
7.0
4.4
Linear Model
(students in 1000’s)
Residual
Square of
Residual
Total Sum of
Squares of
Residuals
The total sum of squares is approximately 0.054 which is less than the total sum of squares
seen with the models presented in #1.
4. Interpret the meaning of the slope and the vertical intercept of the model in terms of enrollment, A ,
and years since 1990, t .
The slope of 0.195 tells us that for each additional year, the number of people in prison, jail, on
probation, or on parole increases by 0.195 (million or 195,000) people. The vertical intercept
tells us that in 1990, the model predicts the number of people in prison, jail, on probation, or
on parole was 4.4 (million or 4,400,000) people.
5. Use the model to predict the number of people in prison, jail, on parole, or on probation in 2003. Is
this an example of interpolation or extrapolation? Why?
y  0.195(13)  4.4  6.935 million people. The regression model predicts that in 2003, there
were 6.935 million (6,935,000) people in prison, jail, on probation, or on parole. This is an
example of interpolation since the prediction is within the given data set.
36
Activity 2.4 #1: Systems of Linear Equations
Activity Design:
Individual or Group
Activity Overview:
Graph, evaluate, and interpret systems of linear equations in graphical form
Advance Preparation: None
Alcohol is metabolized by the body in such a way that the blood alcohol content (BAC) decreases
linearly. Two studies by the National Institute on Alcohol Abuse and Alcoholism were conducted and
showed that:
• for a group of fasting males of moderate weight (150 – 180 pounds) who consumed four drinks
rapidly, the BAC rose to a maximum of 0.08 g/100 mL about an hour after the drinks were
consumed. Three hours later, the BAC had decreased to 0.04 g/100 ML.
• for a second group of fasting males of light weight (100 – 150 pounds) who consumed four
drinks rapidly, the BAC rose to a maximum of 0.09 g/100 mL about an hour after the drinks
were consumed. Three hours later, the BAC had decreased to 0.036 g/100 ML (Source:
www.niaaa.nih.gov ).
1. Construct a formula for the BAC of the males of moderate weight as a function of the time the
maximum BAC (0.08 g/100 mL) was reached, M (t ) .
0.04
t
3
= 0.08 − 0.0133t
M=
(t ) 0.08 −
2. Construct a formula for the BAC of the males of light weight as a function of the time the
maximum BAC (0.08 g/100 mL) was reached, L(t ) .
0.036
t
3
= 0.09 − 0.018t
L=
(t ) 0.09 −
3. Explain the meaning of the slope and vertical intercepts in the real world context for both functions
M (t ) and L(t ) .
For M (t ) the slope means that for the men of moderate weight the BAC was decreasing 0.0133
g/100 mL every hour and the vertical intercepts means that the BAC was initially 0.08 g/100
mL.
For L(t ) the slope means that for the men of light weight the BAC was decreasing 0.018 g/100
mL every hour and the vertical intercepts means that the BAC was initially 0.09 g/100 mL
37
4. Graph M (t ) and L(t ) over a reasonable domain.
5. Using the formulas for M (t ) and L(t ) , determine if the two groups of men had the same BAC over
the time interval from 0 to 8 hours.
M=
(t ) 0.08 − 0.0133t and L=
(t ) 0.09 − 0.018t so we need to first find out when the BAC for
the two groups of males are equal.
M (t ) = L(t )
0.08 − 0.0133t =
0.09 − 0.018t
−0.01 =
−0.0047t
2.12 ≈ t
After about 2.12 hours, the BAC was the same for both groups.
6. Using your answers to problems (5) and (6) above, determine over what time interval the males of
moderate weight had a greater BAC then males of light weight.
We can see that before 2.12 hours when the two groups had the same BAC that the males of
moderate weight had a higher BAC than the males of light weight.
7. Find the horizontal intercepts for each function and explain what they mean in this context.
M=
(t ) 0.08 − 0.0133t
=
0 0.08 − 0.0133t
−0.08 =
−0.0133t
−0.08
=t
−0.0133
6.02 ≈ t
This tells us that after about 6 hours the males of moderate weight will have no alcohol in their
bloodstream.
L=
(t ) 0.09 − 0.018t
=
0 0.09 − 0.018t
−0.09 =
−0.018t
−.09
=t
−0.018t
5=t
This tells us that after about 5 hours the males of light weight will have no alcohol in their
bloodstream.
38
Activity 2.4 #2: Systems of Linear Equations
Activity Design:
Individual or Group
Activity Overview:
Graph, evaluate, and interpret systems of linear equations in graphical form
Advance Preparation: None
Based on data collected by the Energy Information Administration from 2006 to 2010, the supply of
United States’ coal may be modeled by S (t ) =
−16.82t + 1238.74 in millions of short tons where t is
the number of years since 2000 while the consumption of United States’ coal may be modeled by
C (t ) =
−16.55t + 1228.24 in millions of short tons where t is the number of years since 2000 (Source:
www.eia.doe.gov/steo ).
1. Explain the meaning of the slope and vertical intercepts in the real world context for both functions
S (t ) and C (t ) .
For S(t) the slope means that the supply of United States’ coal was decreasing 16.82 million
tons every year and the vertical intercept means that the coal supply was initially 1238.74
million tons.
For C(t) the slope means that the consumption of United States’ coal was decreasing 16.55
million tons every year and the vertical intercept means that the coal consumption was initially
1228.24 million tons.
2. Graph S (t ) and C (t ) over the years 2000 to 2015 and from 1000 to 1250 million short tons of coal.
3. The two graphs appear to be almost parallel from the year 2000 to 2015. Explain why this makes
sense.
It would seem reasonable that whatever amount of coal is supplied would be consumed at near
the same rate. Coal companies are going to try and match their production to consumption as
closely as possible.
39
4. Using the formulas for S (t ) and C (t ) , determine if the two functions ever intersect and if so, explain
what this means.
Since S (t ) =
−16.82t + 1238.74 and C (t ) =
−16.55t + 1228.24
S (t ) = C (t )
−16.82t + 1238.74 =
−16.55t + 1228.24
−0.27t =
−10.47
t ≈ 38.78
In 2039 the coal supply and consumption will be equal.
Use the hypothetical graph below to answer questions 5, 6, and 7. Note that a point of intersection
occurs at (8,1100).
5. Explain what the point of intersection means in this context.
The point of intersection means that in 2008 supply and consumption of coal were equal
in the United States.
6. Explain what the graph tells you for the years 2000 to 2008.
From 2000 to 2008 the supply outpaced the consumption in the United States.
7. Explain what the graph tells you for the years 2008 to 2015.
From 2008 to 2015 the consumption outpaced the supply in the United States.
40
Activity 2.5 #1: Systems of Linear Inequalities
Activity Design:
Group
Activity Overview:
Use a system of equations to determine a mixture
Advance Preparation: none
Many hikers enjoy eating dried fruit while in the back country. The nutritional content of
the dried fruit mix varies depending upon the composition of the mixture. The table
below shows the nutritional content of Trader Joe’s® Organic Banana Chips and of
Trader Joe’s® Omega Cranberry Pieces (sweetened) as printed on the package labeling.
Banana Chips
Cranberry Pieces
Serving size
Carbohydrates
30 g
13 g
(about 13 pieces)
28 g
24 g
(about ¼ cup)
Source: www.nutri-facts.com
Calories
Fat
160
11 g
110
1g
1. Let b be the number of 30 g servings of banana chips and c be the number of 28 g
servings of cranberries in the mixture. Do the following:
a. Write an expression that gives the total amount of carbohydrates in the mixture.
13b + 24c
b. Write an expression that gives the total amount of calories in the mixture.
160b + 110c
c. Write an expression that gives the total amount of fat in the mixture.
11b + 1c
d. Write an expression that gives the total number of grams of ingredients (bananas
and cranberries) in the mixture.
30b + 28c
41
2. A hiker wants to construct a mix of up to 840 grams of dried fruit. The mix must
contain at least 500 grams of carbohydrates, at least 1600 calories, and at most 65
grams of fat. (For a person on a 2000-calorie diet, the government recommends that
fat consumption not exceed 65 grams.) Identify four different mix options that meet
the hiker’s criteria.
We write the inequalities defining the hiker’s criteria.
Ingredients: 30b + 28c ≤ 840
Carbohydrates: 13b + 24c ≥ 500
Calories: 160b + 110c ≥ 1600
Fat: 11b + 1c ≤ 65
The solution region must lie below the ingredients and fat lines and above the
carbohydrate and calories lines. The region bordered above by the ingredients line,
to the right by the fat line, below by the carbohydrates line, and to the left by the
vertical axis is the solution region. (See the figures below).
Four of the possible solutions to the mixture problem are graphed. They include:
1 30-gram serving of bananas and 26 28-gram servings of cranberries
2 30-gram servings of bananas and 22 28-gram servings of cranberries
3 30-gram servings of bananas and 24 28-gram servings of cranberries
4 30-gram servings of bananas and 20 28-gram servings of cranberries
Any point inside the interior of the trapezoidal region is a solution. All points on the
boundary lines that border the interior of the region are also solutions.
42
Activity 2.5 #2: Systems of Linear Inequalities
Activity Design:
Group or Individual
Activity Overview:
Use a system of linear inequalities to find an investment strategy
Advance Preparation: Optional: Bring in current investment performance information
Many financial planners advise their clients to diversify their investments. By putting
their money in a variety of different types of investments, consumers can decrease their
risk and increase their chances of financial success.
As of March 2008, Fidelity Investments advertised the following mutual funds on their
website (www.fidelity.com).
Annualized Returns
1 year
5 year 10 year
Fidelity High
Income Fund
Fidelity Fifty
Fund
1.54%
Share Price
52week low 52week high
9.13%
3.73%
$8.29
$9.20
1.35% 10.52%
7.70%
$17.98
$27.30
One way to measure risk is to determine the volatility of the share price. That is, how
much does the share price fluctuate. One numeric way to calculate volatility is to divide
the difference in the maximum and minimum share prices by the maximum share price.
27.30  17.98
9.20  8.29
Fifty:
 0.10
 0.34
9.20
27.30
This number will always be between 0 and 1. The closer the value is to 1 the more
volatile the share price. The closer the value is to 0 the more stable the share price.
High Income:
1. Gather investment information about two different investment accounts. (You may use
those above or may construct a similar table for your own investment accounts.)
2. Identify variables to represent the amount of money invested in each investment
account. Then do the following:
Let H be the amount invested in the High Income Fund and F be the amount invested
in the Fifty Fund.
a. Write an expression that gives the total amount invested in both accounts.
HF
b. Assuming that the accounts will earn their 10-year annualized return, write an
expression that gives the total amount of money earned on the investments in a year.
0.0373H  0.0770F
43
c. Assuming that the volatility of the accounts will remain constant, we can construct a
measure of the overall volatility of the investment by multiplying the amount
invested in each account by the corresponding volatility number and summing the
results. Write an expression that does this.
0.10H  0.34F
3. An investor has up to $5000 to invest between the two investment accounts. The
investor wants to earn at least $250 annually while keeping the volatility measure at or
below $1000. Identify four different investment options that meet the investor’s criteria.
H  F  5000
0.0373H  0.0770 F  250
0.10 H  0.34 F  1000
 There is up to $5000 to invest
 The annual return is at least $250
 The volatility measure is at most $1000
We graph the boundary lines of the three inequalities. The triangular region at the
center of the figures below satisfies all of the criteria.
Four of the possible solutions to the investment problem are shown. They include:
$1900 in the Fifty Fund and $3000 in the High Income Fund
$2000 in the Fifty Fund and $2900 in the High Income Fund
$2100 in the Fifty Fund and $2600 in the High Income Fund
$2200 in the Fifty Fund and $2300 in the High Income Fund
Any point inside the interior of the triangular region is a solution. All points on the
boundary lines that border the interior of the triangular region are also solutions.
44
Activity 3.1 #1: Horizontal and Vertical Shifts
Individual or Group
Explain what change in a function equation results in a horizontal or vertical
shift
Advance Preparation: None
Activity Design:
Activity Overview:
The table below is data representing the weight of average healthy babies brought to term (born on the
at or near 40 weeks of pregnancy) in the United States from birth to 12 months. (Source:
www.Eirpharm.com ).
Average Baby Weight
Month
(in pounds)
(since birth)
t
A(t )
0
1
2
3
4
5
6
9
12
7.5
8.9
10.7
12.5
14.3
16.0
17.4
19.8
22.1
On January 29, 2007 a baby was born in Cancun, Mexico weighing 14 pounds, 8 ounces (14.5
pounds). The baby was nicknamed by nurses “Super Tonio and was born via Caesarean section to his
mother age 23 and father age 38. Doctors noted that the newborn drinks 5 ounces of milk every three
hours and wears diapers meant for six-month olds. (Source: www.baby2see.com).
1. Based on the average baby weight and Super Tonio’s data above, create a table of values and
draw the associated graph that represents the weight of Super Tonio relative to the average
baby weight. Assume that Super Tonio grows at the same rate as the average baby does after
birth. How do your table and graph compare to the model A(t ) ?
Month
Average Baby
Super Tonio’s Weight
Weight
(since birth)
(in pounds)
(in pounds)
t
S (t )
A(t )
0
1
2
3
4
5
6
9
12
7.5
8.9
10.7
12.5
14.3
16.0
17.4
19.8
22.1
14.5
15.9
17.7
19.5
21.3
23.0
24.4
26.8
29.1
The graph of S(t) is shifted up vertically 7 units and the output values in the table are
increased by 7.
45
2. Write an equation that relates A(t ) and S (t ) .
S (t ) = A(t ) + 7
On February 3rd, 2007 Great Britain’s tiniest baby to survive was born prematurely. Ruby
weighed just 12.5 oz (approximately 0.78 pounds) at her birth but has added weight
consistently and is now a healthy, happy baby. (Source: www.baby2see.com).
3. Suppose that after birth “Little Ruby” grew at the same rate as the average baby. Create a table
showing the weight of “Little Ruby”, R(t ) and her difference from the average weight.
Month
(since birth)
t
0
1
2
3
4
5
6
9
12
Average Baby
Weight
(in pounds)
A(t )
Little Ruby’s Weight
(in pounds)
R(t )
7.5
8.9
10.7
12.5
14.3
16.0
17.4
19.8
22.1
4. Create a graph of R(t ) with respect to A(t ) .
5. Write an equation that relates A(t ) and R(t ) .
R(t ) = A(t ) − 6.72
46
Difference from
Average
Activity 3.1 #2: Horizontal and Vertical Shifts
Individual or Group
Explain what change in a function equation results in a horizontal or vertical
shift
Advance Preparation: None
Activity Design:
Activity Overview:
The table below is data from a study of the carry distance for drives using low (0.5 – 0.75 inches), mid
(1.5 - 1.75 inches), and high (2 – 2.25 inches) tee heights for golfers with low ability levels (handicaps
of 20+). (Source: Golf Magazine, June 2006).
Tee Height
(in inches)
Carry Distance
(in yards)
h
C (h)
0.50
0.75
1.50
1.75
2.00
2.25
160.85
164.38
174.15
176.10
178.24
174.94
For problems 1 – 4, use the table above to complete tables for m , n , o , and p where:
a. m(h) = C (h) − 10
b. n(h) = C (h) + 12
c. o(h) = C (h) − 8
d. p (h) = C (h) + 4
1.
m(h)
h
0.50
0.75
1.50
1.75
2.00
2.25
2.
n(h)
h
0.50
0.75
1.50
1.75
2.00
2.25
47
3.
o(h)
h
0.50
0.75
1.50
1.75
2.00
2.25
4.
p(h)
h
0.50
0.75
1.50
1.75
2.00
2.25
5. Explain how a graph of each function m , n , o , and p relates to the graph of C (h) .
a. m(h) is moved down 10 units
b. n(h) is moved up 12 units
c. o(h) is moved down 8 units
d. p (h) is moved up 4 units
6. Verbally describe what each function m , n , o , and p means in terms of the context.
a. m(h) = C (h) − 10 means a golfer who hits the ball 10 yards shorter than average.
b. n(h) = C (h) + 12 means a golfer who hits the ball 12 yards longer than average.
c. o(h) = C (h) − 8 means a golfer who hits the ball 8 yards shorter than average.
d. p (h) = C (h) + 4 means a golfer who hits the ball 4 yards shorter than average.
7. If q (h) = C (h − 0.50) , evaluate q (2.00) and explain what the value means in this context.
q (2.00) = 174.15 is a player who hits a drive that carries 174.15 yards with a tee of 2 inches
tall.
48
Activity 3.2 #1: Horizontal and Vertical Reflections
Individual or Group
Explain what change in a function equation results in a horizontal or vertical
reflection
Advance Preparation: None
Activity Design:
Activity Overview:
For problems 1 – 4, do the following:
a. Graph f (x) using a graphing calculator or a table of values.
b. Describe the transformations required on f (x) to create g (x) .
c. Write the formula for g (x) as a function of x.
1.
2.
3.
f ( x) = 2 x − 3 and g ( x) = − f (− x)
a.
b. The function f (x) has to be reflected
vertically and horizontally.
c. g ( x) = −(−2 x − 3) = 2 x + 3
f ( x) = 3 x 2 and g ( x) = − f ( x)
a.
b. The function f (x) has to be reflected
vertically.
c. g ( x) = −3 x 2
f ( x) = x and g ( x) = − f (− x)
a.
b. The function f (x) has to be reflected
vertically and horizontally.
c. g ( x) = − − x = − x
4.
f ( x) = x 3 − 2 and g ( x) = f (− x) − 2
a.
b. The function f (x) has to be reflected
horizontally and shifted down two units.
c. g ( x) = (− x)3 − 2 − 2 = − x 3 − 4
49
In Exercises 5 – 10, use the table of values to evaluate each expression.
f (x)
24
18
11
3
-6
-16
-27
-39
-55
x
-6
-4
-2
0
2
4
6
8
10
5. − f (x) when x = 6 .
27
6. − f (x) when x = −4 .
-18
7.
f (− x) when x = 4 .
18
8.
f (− x) when x = −8 .
-39
9. − f ( x + 4) − 9 when x = 2 .
27-9=18
10. − f (− x) + 3 when x = 0 .
0
50
Activity 3.2 #2: Horizontal and Vertical Reflections
Individual or Group
Explain what change in a function equation results in a horizontal or vertical
reflection
Advance Preparation: None
Activity Design:
Activity Overview:
For problems 1 – 5, do the following:
a. Graph f (x) using a graphing calculator or a table of values.
b. Describe the transformations required on f to create g.
c. Write the formula for g in terms of x.
1.
2.
3.
4.
f ( x) = x + 5 and g ( x) = − f (− x) .
a.
b. The function f (x) has to be reflected
vertically and horizontally.
c. g ( x) = −(− x + 5) = x − 5
f ( x) = − x and g ( x) = − f (− x) .
a.
b. The function f (x) has to be reflected
vertically and horizontally.
c. g ( x) = − x
f ( x) = x + 7 and g ( x) = − f ( x) .
a.
b. The function f (x) has to be reflected
vertically.
c. g ( x) = −( x + 7) = − x − 7
f ( x) = x 3 − 12 and g ( x) = − f ( x) + 4 .
a.
b. The function f (x) has to be reflected
vertically and shifted up 4 units.
c. g ( x) = − x 3 + 12 + 4 = − x 3 + 16
51
5.
f ( x) = −3 x − 2 and g ( x) = − f (− x) + 3 .
a.
b. The function f (x) has to be reflected
vertically and horizontally and shifted up
3 units.
c. g ( x) = −3 x + 5
In Exercises 6 – 10, use the table of values to evaluate each expression.
f (x)
34
28
21
13
4
-6
-17
-29
-45
x
-9
-7
-5
-3
-1
1
3
5
7
6.
− f (x) when x = 3 .
17
7.
f (− x) when x = 1 .
4
8.
f (− x) when x = −5 .
-29
9. − f ( x + 4) − 9 when x = −1 .
8
10. − f (− x) + 3 when x = −3 .
20
52
Activity 3.3 #1: Vertical Stretches and Compressions
Individual or Group
Explain what change in a function equation results in a vertical compression or
stretch
Advance Preparation: None
Activity Design:
Activity Overview:
Stereo amplifiers can take weak signals from a radio and transform them into stronger signals to power
a set of speakers. The graphs below show the strength of a radio signal of 105.9 MHz (megahertz) in
1 th
volts, r , as a function of the time (in
of a second), t before amplification and the function
105.9
a (t ) after the signal has been amplified.
1. Describe what type of transformation converts r (t ) into a (t ) .
The graph was vertically stretched by a factor of 10.
2. The voltage gain of an amplifier refers to the factor by which a radio signal is amplified.
Considering the graph above, what is the voltage gain of this amplifier? (Note: the
amplification that actually takes place is much larger in reality but in order to see the
amplification graphically the factor must remain relatively small for this activity.)
The voltage gain is 10.
3. If the original radio signal is modeled by the function r (t ) , give a formula for the amplified
radio signal a (t ) as a transformed function of r (t ) .
a (t ) = 10r (t )
4. Explain why the t - intercepts of both functions remain the same after the vertical stretch has
occurred.
When the t - intercepts are multiplied by the factor of 10 they do not change because
their output value is 0 and 10 ⋅ 0 = 0 .
53
For problems 5 and 6, consider the table below that gives the values of a radio signal in volts, v being
received by a receiver as a function of time, t .
Time
1 th
(in
of a second)
99.9
t
99.9 MHz
Radio Signal
(in volts)
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
v(t )
-2
0
2
0
-2
0
2
0
-2
5. Describe the function v(t ) in words. What do these data indicate about the radio signal’s
voltage?
The function oscillates between -2 and 2 volts over the time of 1/99th of a second.
6. Make a table of values for g (t ) = 125v(t ) . What voltage gain would an amplifier have to make
this type of transformation on the radio wave?
Time
1 th
(in
of a second)
99.9
t
Amplified 99.9 MHz
Radio Signal
(in volts)
g (t )
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
The voltage gain would be 125.
54
Activity 3.3 #2: Vertical Stretches and Compressions
Individual or Group
Explain what change in a function equation results in a vertical compression or
stretch
Advance Preparation: None
Activity Design:
Activity Overview:
Using the graph of f below, match the functions (1) – (5) with a graph (a) – (i). (Note: Some graphs
will not be used.)
1. =
y 3 f ( x) + 1
1
2.
y = f ( x)
4
3.
y = − f ( x) + 2
4.
y = f (− x)
5.
y = f ( x + 2) + 1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
1-d; 2-i; 3-c; 4-a; 5-g
55
6. The number of gallons of gasoline, g (t ) , in a vehicle’s tank is a function of the time in hours.
Match each story to one expression.
a. I added 1.5 extra gallons to the amount in the tank.
b. I added enough gas to my tank to increase it by 1.5 times as much.
c. I am wondering how much gas was in the tank 1.5 hours ago.
i.
ii.
iii.
1.5 g (t )
g (t + 1.5)
g (t ) + 1.5
a. iii.
b. i.
c. ii.
7. The U.S. population in millions can be modeled by the function P(t ) as a function of time t , in
years. Match each statement a – d with one of the formulas i – viii.
a. The population 5 years before today.
b. Today’s population plus 5 million immigrants
c. Fifty percent of the population we have today.
d. The population after 500,000 people have left the country.
i.
ii.
iii.
iv.
v.
P(t ) − 5
P(t − 5)
0.5 P (t )
P(t ) + 5
P(t + 5)
P(t )
vi.
0.5
vii. P(t ) + 0.5
viii. P(t ) − 0.5
a.
b.
c.
d.
i.
iv.
iii.
viii.
8. Let B(t ) be the number of birds living in a field in month t . What do the following expressions
represent?
a. B(t + 1)
b. 2 B(t )
B(t + 1) is the number of birds living in the field one month after the initial
time of measurement
2 B (t ) is twice the number of birds living in the field at time t.
56
Activity 3.4 #1: Horizontal Stretches and Compressions
Individual or Group
Explain what change in a function equation results in a horizontal compression
or stretch
Advance Preparation: None
Activity Design:
Activity Overview:
Lighthouses are used to warn ships at sea of impending danger due to jagged, rough shorelines.
The beacon of light emitted by a lighthouse turns one time per minute, and its beam sweeps across
a ship’s path. The graph of the function I (t ) shows the intensity (brightness) of the light striking
the ship as a function of time in minutes. (Source: www.usenix.org)
1. Suppose that the lighthouse beacon is changed to turn twice as fast as before, so that its beam
sweeps past the ship twice instead of once each minute. Describe the transformed
function B(t ) .
The function will have the same outputs except that within one minute there will be two
peak intensities.
2. Write the function B(t ) as a transformed function in terms of I (t ) .
B (t ) = I (2t )
1
of the original rate, so
3
that its beam sweeps past the ship once every three minutes instead of once per minute.
Describe the new function N (t ) that models the light intensity upon the ship.
3. Now suppose that the lighthouse’s beacon is slowed so that it turns at
The function will have the same outputs except that a peak intensity will occur after
three minutes.
4. Describe the change in the graph of N (t ) with respect to I (t ) .
The graph is horizontally stretched by a factor of 3.
57
5. Values of the function f (x) are in the table below. Make a table of the function
1
g ( x) = f ( x) .
2
f (x)
x
-3
-2
-1
0
1
2
3
0
4
0
-2
0
-2
2
x
g (x)
6. Use the table from part 5 to make a table of the function h( x) = f (3 x) .
x
h(x)
58
Activity 3.4 #2: Horizontal Stretches and Compressions
Individual or Group
Explain what change in a function equation results in a horizontal compression
or stretch
Advance Preparation: None
Activity Design:
Activity Overview:
As a paddlewheel turns driving a boat downriver, each paddle blade moves in such a way that its
distance in feet, d , from the water’s surface after t seconds is shown by the graph of d (t ) shown
below.
1. Suppose that the paddlewheel begins to turn twice as fast as before. Graph the transformed
function n(t ) over the first 30 seconds.
2. Write the function n(t ) as a transformed function in terms of d (t ) .
n(t ) = d (2t )
1
3. Now suppose that the paddlewheel is slowed so that it turns at of the original rate. Graph the
2
new function k (t ) that models the distance of the paddlewheel above the water’s surface over
the first 30 seconds.
4. Describe the change in the graph of k (t ) with respect to d (t ) .
The graph is horizontally stretched by a factor of 2.
59
1
5. Values of the function f (x) are in the table below. Make a table of the function g ( x) = f ( x) .
3
f (x)
x
-6
-4
-2
0
2
4
6
0
4
0
-2
0
-2
2
x
g (x)
6. Use the table from (5) to make a table of the function h( x) = f (1.5 x) .
x
h(x)
60
Activity 4.1 #1: Variable Rates of Change
Activity Design:
Individual or Group
Activity Overview:
Determine the rate of change of a function over given intervals, determine
concavity and increasing/decreasing behavior of functions.
Advance Preparation: None
The table below is data modeled of the batted ball speed (in MPH) for various bat weights (in ounces)
for Major League Baseball players in 2007. (Source: Popular Mechanics, June 2007).
Major League Baseball Batted Ball
Speed
B
w
Batted Ball Speed
(miles per hour)
Bat Weight
(ounces)
20
25
30
35
40
45
50
68.12
72.75
76.10
78.16
78.94
78.43
76.64
1. Determine if the function B( w) has a constant or variable rate of change over the interval of years
provided in the table.
Major League Baseball Batted Ball Speed
w
Bat Weight
(ounces)
B
Batted Ball Speed
(miles per hour)
20
68.12
25
72.75
30
76.10
35
78.16
40
78.94
45
78.43
50
76.64
61
Rate of change
2. Give the interval of bat weights over which the function B ( w) is increasing.
The interval of bat weights that B( w) is increasing from 20 to 40 ounces because the rate is
positive.
3. Give the interval of bat weights over which the function B ( w) is decreasing.
The interval of bat weights that B( w) is decreasing is from 40 to 50 ounces because the rate is
negative.
4. Give the interval of bat weights that the function B( w) is concave up.
B ( w) is not concave up over the interval of bat weights from 20 to 50 ounces because the
rates are all decreasing.
5. Give the interval of bat weights that the function B( w) is concave down. Explain what the concavity
tells about the batted ball speed.
B ( w) is concave down over the interval of bat weights from 20 to 50 ounces because the rates
are all decreasing. This means that the rate at which the batted ball speed is changing is
decreasing.
6. B( w) can be modeled by the formula B ( w) =
−0.0257 w2 + 2.0829 w + 36.7429 . Using this formula
for B( w) , what is the projected maximum batted ball speed? Does this answer make sense in the
context of the problem?
The maximum batted ball speed using B( w) is w = 40.5 is 78.9 miles per hour. This answer
makes sense because as the bat gets heavier the player is unable to swing it as fast.
7. Using B ( w) =
−0.0257 w2 + 2.0829 w + 36.7429 , compute the average rate of change for the weights
w = 23 and w = 38 and explain what this number means in the real-world context.
The average rate of change is computed
78.782 − 71.054 miles per hour
miles per hour
.
= 0.5152
38 − 23 ounces
ounces
This tells us that if the bat weight increases from 23 to 38 ounces the batted ball speed
increases on average 0.5152 miles per hour per ounce.
8. Using B ( w) =
−0.0257 w2 + 2.0829 w + 36.7429 , estimate the instantaneous rate of change for a bat
weight of 43 ounces explain what this number means in the real-world context.
The instantaneous rate of change is computed by taking values for bat weights close to 43
ounces and finding the rate of change. We choose weights 42 and 43.
78.788 − 78.890 miles per hour
miles per hour
.
= −0.102
43 − 42 ounces
ounces
This tells us that for a bat that is 43 ounces the rate of change in batted ball speed decreases
0.102 miles per hour for each additional ounce.
62
Activity 4.1 #2: Variable Rates of Change
Activity Design:
Individual or Group
Activity Overview:
Determine the rate of change of a function over given intervals, determine
concavity and increasing/decreasing behavior of functions.
Advance Preparation: None
The table below is data modeled after the average retail gas prices per gallon (in dollars) for regular
gasoline for years since 2000. (Source: Modeled from the Energy Information Administration,
http://tonto.eia.doe.gov/dnav/pet/hist/mg_rt_usA.htm ).
Retail Price Per Gallon
P
Average price per
gallon of regular
gasoline
(dollars)
1.41
1.41
1.48
1.63
1.84
2.12
2.49
2.92
y
Year
0
1
2
3
4
5
6
7
1. Determine if the function P ( y ) has a constant or variable rate of change over the interval of years
provided in the table.
Retail Price Per Gallon
P
Year
Average price per
gallon of regular
gasoline
(dollars)
0
1.41
1
1.41
2
1.48
3
1.63
4
1.84
5
2.12
6
2.49
7
2.92
y
63
Rate of change
2. Give the interval of years that the function P ( y ) is increasing.
The interval of years that P ( y ) is increasing is from years 1 to 7 because the rate of change is
positive.
3. Give the interval of years that the function P ( y ) is decreasing.
P ( y ) does not decrease over the interval of years from 0 to 7 because the rate of change is
positive.
4. Give the interval of years that the function P ( y ) is concave up. Explain what the concavity tells
about the average price per gallon.
P ( y ) is concave up over the interval of years from 0 to 7 because the rates of change are
increasing.
5. Give the interval of years that the function P ( y ) is concave down.
P ( y ) is not concave down over the interval of years from 0 to 7 because the rates are
increasing.
6. P ( y ) can be modeled by the formula P ( y ) = 0.0360 y 2 − 0.0369 y + 1.4125 . Using this formula
for P ( y ) , what is the projected minimum average price per gallon? Does this answer make sense in
the context of the problem?
The minimum average price per gallon using P ( y ) occurs in y = 0 .512 and is $1.40. This
answer makes sense as the cost of gasoline has continued to rise during the decade.
7. Using P ( y ) = 0.0360 y 2 − 0.0369 y + 1.4125 , compute the average rate of change for the years
y = 2 and y = 6 and explain what this number means in the real-world context.
The average rate of change is computed
$2.49 − $1.48
dollars
.
= 0.25
6 − 2 years
year
This tells us that the average cost of a gallon of gasoline increased by $0.25 per year.
8. Using P ( y ) = 0.0360 y 2 − 0.0369 y + 1.4125 , estimate the instantaneous rate of change for the year
10 and explain what this number means in the real-world context.
The instantaneous rate of change is estimated by taking the years close to 10 and finding the
rate of change. We choose years 9 and 10.
$4.60 − $3.96 $0.64
.
=
10 − 9 years
year
This tells us that during the year 2010 the average price per gallon will rise $0.64.
64
Activity 4.2 #1: Modeling with Quadratic Functions
Activity Design:
Activity Overview:
Individual or Group
Create quadratic functions to model situations, describe relationships, and
make predictions.
Advance Preparation: None
The following data comes from Crime in the United States 2000, Uniform Crime Report, FBI.
Cumulative Homicides Due to a
Romantic Triangle
R
t
Homicides between
Years
the Start of 1991 and
(Since 1990) the End of the Year
Shown
1
314
2
648
3
1,088
4
1,459
5
1,739
6
1,928
7
2,104
8
2,291
9
2,428
10
2,550
1. Use regression to find the quadratic function, R(t ) , that best models the data set.
R(t )  20.455t 2  471.81t  152.53
2. Using the model, predict the function value when t  11 . Explain what this value means.
R(11)  20.455 11  471.8111  152.53  2562.3
2
In 2001, that is when t  11 , we predict that the cumulative homicides due to a romantic
triangle is about 2562 homicides.
65
3. Using the model R(t ) , explain the practical meaning of the parameters a, b, and c of the quadratic
function model.
R(t )  20.455t 2  471.81t  152.53
Initially (in 1990), the model indicates that the number of homicides due to a romantic triangle
was, hypothetically, –152.53. We acknowledge that this value does not make sense in the
context of the situation, but is the vertical intercept. This value initially increases by 471.81
homicides per year. This rate of change is not constant but decreases at a rate of 40.91
homicides per year for each additional year.
4. Solve the equation R(t )  1000 graphically on the interval 0  t  10 . Explain what the solution to
the equation means in the context of the situation.
We will estimate the solution to the equation graphically.
We estimate that when t  3 , R(t )  1000 . This tells us that there were a total of 1000
cumulative homicides due to a romantic triangle from the start of 1991 to the end of about
1993.
5. Using the model R(t ) , show that the second differences are constant. Provide an explanation for this
value in the context of the situation.
t
R(t )  20.455t 2  471.81t  152.53
1
2
3
4
5
6
7
298.83
709.27
1078.81
1407.43
1695.15
1941.95
2147.85
first
second
differences differences
The common second difference describes the rate of change of the rate of change. The rate of
change decreases at a rate of about 41 homicides per year for each additional year.
66
Activity 4.2 #2: Modeling with Quadratic Functions
Activity Design:
Activity Overview:
Individual or Group
Create quadratic functions to model situations, describe relationships, and
make predictions.
Advance Preparation: None
The data in the table show the sales, in millions of dollars, of electronic gaming software (Source:
Statistical Abstract of the United States, 2008; Table 1000).
Electronic Gaming Software
Factory Sales
t
S
Years
Sales
(Since 2000) (millions of dollars)
3
4
5
6
7
7065
7350
8041
9006
10087
1. Use regression to find the quadratic function, S (t ) , that best models the data set.
S (t )  133.29t 2  562.86t  7525.4
2. Using the model predict the function value when t  10 . Explain what this value means.
S (10)  133.2910  562.8610  7525.4  15226
In 2010, that is when t  10 , we predict that the total sales of electronic gaming software will
be $15,226 million.
2
3. Using the model S (t ) , explain the practical meaning of the parameters a, b, and c of the quadratic
function model.
S (t )  133.29t 2  562.86t  7525.4
Initially (in 2000), the model indicates that the sales of electronic gaming software was about
$7525 million. This value initially decreases by $562.86 per year. This rate of change is not
constant but increases at a rate of $266.58 per year for each additional year.
67
4. Solve the equation S (t )  10,000 graphically. Explain what the solution to the equation means in the
context of the situation.
We estimate the solution to the equation graphically.
We estimate that when t  7 , S (t )  10,000 . This tells us that there were about $10,000 million
(about $10 billion) in electronic gaming factory sales at the end of 2007.
5. Using the model S2 (t ) , show that the second differences are constant. (Any variations in value are
due to rounding.) Provide an explanation for the second difference in the context of the situation.
t
S (t )  133.29t 2  562.86t  7525.4
0
1
2
3
4
5
6
7525.4
7095.8
6932.8
7036.4
7406.6
8043.4
8946.7
first
second
differences differences
The common second difference describes the rate of change of the rate of change. The rate of
change increases at a rate of about $266.6 million per year for each additional year.
68
Activity 4.3 #1: Quadratic Function Graphs and Forms
Individual or Group
Express quadratic relationships in multiple forms and solve real-world
problems using the quadratic formula.
Advance Preparation: None
Activity Design:
Activity Overview:
When determining the price of a ticket for an event (sporting event, music concert, play, or training
event), market research is conducted to determine the optimum ticket price. That is, what price will
maximize the revenue generated from ticket sales? Suppose that for an upcoming concert at a small
town theater, it is determined from past experience that tickets sold for $75 each will result in 1000
people in attendance. Further, suppose that market research indicates that for each additional $5
increase in ticket price, 20 fewer tickets will be sold.
1. Complete the following table.
Number of $5 price increases
0
5
10
15
20
25
Ticket price
$75
Number of tickets sold
Revenue
2. Describe the pattern you see in the table. Specifically, as the number of $5 ticket price increases
goes up, what happens to the total revenue?
Initially, increasing the ticket price results in greater revenue since the smaller number of
tickets sold is compensated by the increase in price. However, the increase in revenue becomes
less and less as the ticket price increases. According to the table, the revenue reaches $105,000
before beginning to decrease.
3. We will let x represent the number of $5 ticket price increases. Write a function describing the cost,
C , of a ticket as a function of x .
C ( x) = 75 + 5 x
4. Again, we will let x represent the number of $5 ticket price increases. Write a function describing
the number of tickets sold, N , as a function of x .
N ( x) = 1000 − 20 x
69
5. Using your work in #3 and #4, write the total revenue, R , as a function of x .
R( x) = C ( x) ⋅ N ( x)
= (75 + 5 x)(1000 − 20 x)
6. Express the function R (x) in standard form. Using the vertex formula from Section 4.3, calculate
the ordered pair representing the vertex of this function. Explain what this vertex means in the
context of the situation.
−3500
R( x) =
(75 + 5 x)(1000 − 20 x)
x=
2(−100)
=
−100 x 2 + 3500 x + 75000
= 17.5
R(17.5) =
−100 (17.5 ) + 3500 (17.5 ) + 75000
2
= $105, 625
The vertex, (17.5,105,625) , tells us that the maximum revenue is reached when 17.5 ticket price
increases of $5 are made and that the revenue is $105,625.
7. Use a graph of R (x) to confirm your work in #6.
8. Solve the equation R( x) = 0 using the quadratic formula. Interpret the meaning of your result.
R ( x) = −100 x 2 + 3500 x + 75000 = 0
x=
Taking only the result that makes sense in
the practical domain, we find that when
x = 50 , R( x) = 0 . This tells us that if 50,
$5 price increases are made (that is,
tickets cost $325), no tickets will be sold
(50 increases times 20 less people per
increase is 1000 fewer people). If no
tickets are sold, no revenue can be made.
− 3500 ± 3500 − 4(− 100 )(75000 )
2(−100)
2
− 3500 ± 12250000 + 30000000
− 200
− 3500 ± 42250000
=
− 200
− 3500 ± 6500 3000 − 10000
=
=
,
− 200
− 200 − 200
= −15,50
=
70
Activity 4.3 #2: Quadratic Function Graphs and Forms
Activity Design:
Activity Overview:
Individual or Group
Express quadratic relationships in multiple forms and solve real-world
problems using the quadratic formula.
Advance Preparation: None
Matthew recently purchases a bow and arrow. To test the maximum distance he could get the arrow to
fly, Matthew pulled back with all his strength, aimed up at angle of about 45o and let the arrow fly. The
total time of the flight of the arrow was 5 seconds. Assume air resistance was negligible and that the
arrow landed at ground level. Hint: The height above the ground of the arrow t seconds after being
shot into the air may be modeled by h(t )  16t 2  vt  k feet where v is the initial velocity (in feet per
second) at which the arrow was shot and k is the initial height (in feet) at the moment of launch.
Because of Matthew’s height, which is known, we will say k  4 feet.
1. Use the fact that the arrow hit the ground after 5 seconds to find the value of v and thus, determine
the function h(t ) .
h(t )  16t 2  vt  4
0  16  5   v  5   4
2
0  400  5v  4
396  5v
396
 79.2 feet per second
5
h(t )  16t 2  79.2t  4
v
2. Write a quadratic function, H (t ) , in vertex form that represents the height of the arrow, H , as a
function of time, t , in seconds.
H (t )  16t 2  79.2t  4
 16  t 2  4.95t   4
 16  t 2  4.95t  6.125625  4  98.01
 16  t  2.475   102.01
2
3. Compared to the function y  x 2 , describe the behavior of the function, H ( x) , you wrote in #2.
The –16 has the effect of reflecting the function y  x 2 across the x -axis and vertically
stretching it by a factor of 16. The 2.475 shifts the function to the right 2.475 units and the
102.01 shifts the function up 102.01 units when compared to y  x 2 .
71
4. Write the vertex of the function, H (t ) , from #2 as an ordered pair. Explain what the x -value
represents. Explain what the y -value represents.
The vertex is  2.475,102.01 . This means that the maximum height reached by the arrow was
102.01 feet after 2.475 seconds.
5. Evaluate H (4) and explain what the result means in the context of this situation.
H (4)  16  4  2.475  102.01  64.8
2
This means that after 4 seconds, the arrow is at a height of 64.8 feet above the ground.
6. Solve the equation h(t )  50 using the quadratic formula. Explain the meaning of the solution(s) to
this equation in the context of the problem situation.
16t 2  79.2t  4  50
16t 2  79.2t  46  0
t
79.2  79.22  4  16  46 
2  16 

79.2  6272.64  2944
32

79.2  3328.64
32
79.2  3328.64 79.2  3328.64
,
32
32
 0.672, 4.28

The arrow reaches a height of 50 feet after 0.672 seconds (on its way up) and again after 4.28
seconds (on its way down).
72
Activity 5.1 #1: Higher-Order Polynomial Function Modeling
Activity Design:
Activity Overview:
Individual or Group
Use the language of rate of change to describe the behavior of a higher-polynomial
function and model with higher-order polynomials.
Advance Preparation: None
The per capita consumption of breakfast cereals (ready to eat and ready to cook) since 1980 is given in the
table.
t
C
Years
Consumption
(Since 1980)
(pounds)
0
12
2
11.9
4
12.5
6
13.1
8
14.2
10
15.4
12
16.6
14
17.4
16
16.6
18
15.6
Source: Statistical Abstract of the United States, 2001; Table 202.
1. Make a scatter plot of these data and, using the idea of rate of change, explain why a cubic polynomial
function best models the data. Compute the model, C (t ) , and round your parameters to three decimal
places.
It appears as though the per capita consumption (after a
brief decline early on) increases at an increasing rate
(concave up) until about 1988. Then, the per capita
consumption increases at a decreasing rate (concave
down) until 1994 where it begins to decrease, but remains
concave down. Because the graph appears to change
concavity once, a cubic model would be appropriate.
C (t )  0.005t 3  0.123t 2  0.376t  12.112
2. Compute C (20) and explain what the result means in the context of these data.
C (20)  0.005  20   0.123  20   0.376  20   12.112
3
2
 13.792 pounds
The model predicts that in 2000 the per capita consumption of breakfast cereal is 13.792 pounds.
73
3. Solve the equation C (t )  16 for t . Explain what the result means in the context of these data.
We use a graph to solve the equation. C (t )  16 when t  10.615 and again when t  18.044 . This
tells us that at both 11 (around 1991) and 18 (1998) years after 1980, the per capita consumption of
breakfast cereal was 16 pounds per year.
4. Calculate any local extreme values and explain what they mean in the context of the situation.
The local minimum is (1.706,11.804) . The local maximum is (14.694,17.281) . The local minimum
value tells us that in sometime in 1982, the per capita consumption of breakfast cereal reached a
minimum value of 11.804 pounds. The local maximum tells us that sometime in 1995, the per capita
consumption of breakfast cereal reached a maximum value of 17.281 pounds.
5. Describe the end behavior for the function model and discuss whether or not you think this end behavior
will accurately predict future consumption of breakfast cereal.
As t   , C (t )   and as t  , C (t )   . It is not reasonable to use this model to
accurately make long term predictions. As we go further and further back in time from t  0 or 1980,
(or before 1980), the per capita consumption of breakfast cereal goes off to infinity. It is unlikely for
there to be an unlimited consumption of cereal prior to 1980. It is also unlikely for the consumption
of cereal to be zero or below zero after about 2001.
6. Consider the equation C (t )  0 . Provide a mathematical argument for the solution to this equation. You
may use the function, its graph, and/or a table in your argument. Also provide an argument as to why this
equation may have no real-world meaning.
Using the graph of C (t ) , we predict that C (t )  0 when t  25 . That is in 2005, 25 years after 1980,
the per capita consumption of cereal is zero. It is very unlikely for the per capita consumption of
cereal ever to be zero since cereal will always be consumed .
74
Activity 5.1 #2: Higher-Order Polynomial Function Modeling
Activity Design:
Activity Overview:
Individual or Group
Use the language of rate of change to describe the behavior of a higher-polynomial
function and model with higher-order polynomials.
Advance Preparation: None
Based on data from 1990 to 2005, the U.S. exports of frozen and canned vegetables are provided in the table
(Source: Statistical Abstract of the United States: 2007, Table 819).
t
Years
(since
1990)
0
5
10
12
13
14
15
V
Exports of
vegetables
(in thousand
metric tons)
529
892
1112
1065
1010
1048
1083
1. Make a scatter plot of these data and, using the idea of rate of change, explain why a quartic polynomial
function best models the data. Compute the model, V (t ) , and round your parameters to three decimal
places.
It appears as though the export amounts increases at a
decreasing rate (concave down) until about 1999. Then,
the export amounts decrease, concave down briefly before
increasing again. The function is concave up while
changing from decreasing to increasing, and then
continues to increase, but at a decreasing rate (concave
down). Because the graph appears to change concavity
twice, a quartic model would be appropriate.
V (t )  0.166t 4  5.112t 3  44.837t 2  44.493t  528.991
2. Compute V (18) and explain what the result means in the context of these data.
V (18)  0.166 18  5.112 18   44.837 18   44.493 18   528.991
4
3
2
 1868.137
The model predicts that in 2008, approximately 1868.137 metric tons of vegetables were exported
from the United States.
75
3. Solve the equation V (t )  1000 for t . Explain what the result means in the context of these data.
We use a graph to solve the equation. V (t )  1000 when t  6.155 . This tells us that about 6.155
years after 1980 or 1987the U.S. exported 1000 metric tons of vegetables.
4. Calculate any local extreme values and explain what they mean in the context of the situation.
One local minimum is (0.546,517.247) and the other is (13.371,1035.835) . The local maximum
is (9.180,1123.223) . The first local minimum value tells us that sometime in 1991, the number of
vegetable exports reached a minimum value of about 517 metric tons. The other local minimum tells
us that sometime in 2004 the number of vegetable exports reached a minimum of about 1036 metric
tons. The local maximum tells us that sometime in 2000, the number of vegetable exports reached a
maximum of about 1123 metric tons.
5. Describe the end behavior for the function model and discuss whether or not you think this end behavior
will accurately predict future amounts of vegetable exports.
As t   , V (t )   and as t  ,V (t )   . It is not reasonable to use this model to accurately
make long term predictions. As we go further and further back in time from t  0 or 1990 (or before
1990), the number of vegetable exports goes off to infinity. It is impossible for there to be an
unlimited number of vegetable exports prior to 1990. Similarly, it is impossible that the number of
vegetable exports will go off to infinity after 2006.
6. Consider the equation V (t )  0 . Provide a mathematical argument explaining why this equation has no
solution. You may use the function, its graph, and/or a table in your argument. Also provide an argument
as to why this equation has no solution based on real-world meaning.
By inspecting the graph, we see that the function V (t ) has function values greater than 0 for all
values of t . Further, we can analyze the function definition,
V (t )  0.166t 4  5.112t 3  44.837t 2  44.493t  528.991 , as done earlier and conclude that based on
the end behavior and the local minimum values that V (t )  0 for all values of t . In the real world,
there will always be exports of vegetables, which substantiates that V (t )  0 for all values of t.
76
Activity 5.2 #1: Power Functions
Individual or Group
Use power functions to model real-life situations and to describe relationships
between quantities using the language of rate of change.
Advance Preparation: None
Activity Design:
Activity Overview:
The speeds at which a landing airplane will hydroplane are dependent upon the air pressure in the tires. The
data show the speeds that hydroplaning may occur at different tire pressure (Source: www.flightsafety.org).
Tire
Pressure
Speed
S
P
(MPH)
(PSI)
25
50
75
100
125
150
45.000
63.640
77.942
90.000
100.620
110.230
1. Complete the table by computing the rate of change between consecutive data points.
Tire Pressure
P (PSI)
Speed
S (MPH)
25
45.000
50
63.640
75
77.942
100
90.000
125
100.620
150
110.230
Average Rate of Change
(MPH/PSI)
2. Examine the average rate of change between data points and explain why a power function might best
model the situation.
As the tire pressure increases, the speed at which the plane will hydroplane increases but by less and
less. We see a relatively high rate of change for lower tire pressures. The rate of change decreases,
becoming closer and closer to 0, as the tire pressure increases.. This is characteristic of a power
function of the form y = axb where 0 < b < 1 .
77
3. Find a power function, S = f (P) , that models these data. Round all parameters to three decimal places.
S = f ( P) = 9 P 0.5
4. Using the language of rate of change, describe the relationship between tire pressure and speed at which
an air plane hydroplanes as given by the power function model.
The power function S = f ( P) = 9 P 0.5 , models the speed at which an airplane will hydroplane given
its tire pressure It tells us that after the initial speed the airplane begins to hydroplane, 9 mph, the
hydroplaning speed increases relatively quickly for every 1 PSI increase in tire pressure. As the tire
pressure increases, the speed increases by an ever decreasing rate. The exponent, b = 0.5 , tells us
that the initial increase will indeed be relatively large and the graph will level off as the tire pressure
increases. This is because the value of b is between 0 and 1.
5. Create a graphical representation of your response for #4.
For each increase in 40 PSI, we see the vertical change
become less and less.
6. Use the power function model to determine the speed at which an airplane will hydroplane if the tire
pressure is 90 PSI.
=
S f=
(90) 9(90)0.5 ≈ 85.381
According to the model, if the tire pressure is 90 PSI the airplane will hydroplane at a speed of about
85 MPH.
7. Write and solve an equation that would determine the tire pressure of an airplane that hydroplanes at a
speed of 70 MPH.
=
S f=
( P) 9 P()0.5
70 = 9 P 0.5
70
= P 0.5
9
2
 70 
=
P   ≈ 60.494
 9 
According to the model, an airplane that hydroplanes at a speed of 70 MPH has tires with a pressure
of 60.494 PSI.
78
Activity 5.2 #2: Power Functions
Individual or Group
Use power functions to model real-life situations and to describe relationships
between quantities using the language of rate of change.
Advance Preparation: Consider having students collect the data using jumbo paperclips.
Activity Design:
Activity Overview:
Stress tests on paperclips can be performed by selecting an angle at which you can repeatedly bend the inner
loop of the paperclip until it breaks. The stress test can be repeated on a selected number of the paper clips,
using a different angle for each one. The following data have been collected in a prior experiment.
Angle of
Bend
Number of
Bends
a
45
90
180
B
51
5
4
2
2
3
1
1
2
270
360
The data point (45, 51) means that if the inner
loop of a paper clip is bent at an angle of 45
degrees, returned to its original position, and
then bent again, it will break after 51 iterations.
1. Examine the average rate of change between data points and explain why a power function might best
model the situation.
As the angle of the bend increases, the number of bends needed to break the paperclip decreases but
by less and less. This is characteristic of a power function of the form y = axb where b < 0 .
2. Find a power function, B = f (a ) , that models these data. Round all parameters to three decimal places.
=
B f=
(a ) 9076.033a −1.491
3. Using the language of rate of change, describe the relationship between the angle of bend and the number
of bends needed to break the paperclip as given by the power function model.
The power function
=
B f=
(a ) 9076.033a −1.491 , which models the number of bends needed to break
the paperclip given the angle of the bend, tells us that initially, the number of bends needed to break
the paperclip decreases relatively quickly for every increase of 1 degree in angle of bend. As the
angle of the bend increases, the number of bends before breaking decreases at an ever decreasing
rate. The exponent, b = −1.491 , tells us that the initial decrease will indeed be relatively large and
the graph will level off as the angle of the bend increases. This is because the value of b is less than
0.
79
4. Assuming that under normal conditions, a paperclip is bent at an angle of 5 degrees, how many times can
the paper clip be bent before breaking?
−1.491
=
B f=
(5) 9076.033 ( 5 )
B ≈ 823.626
According to the model, if the angle of the bend is 5o, the paperclip will bend after about 824
consecutive bends.
5. Write and solve an equation that would determine the angle at which the paperclip is bent if it broke after
100 consecutive bends.
=
B f=
(a ) 9076.033 ( a )
100 = 9076.033 ( a )
−1.491
−1.491
100
1
=
9076.033 ( a )1.491
=
(a)
1.491
9076.033
= 90.76033
100
a = 1.491 ( 90.76033)
a ≈ 20.566
According to the model, a paperclip will break after about 100 bends if bent at an angle of about
20.57 degrees.
6. Explain what should happen, in the context of the situation, to the dependent variable as the independent
variable approaches 0. Use graphs and words to support your explanation.
As the angle of the bend (independent variable) approaches 0, the
number of bends needed to break the paperclip (dependent variable)
goes to infinity. This makes sense since if the paperclip is not bent
(angle is 0 degrees), it will never break (number of bends goes to
infinity).
7. Explain what should happen, in the context of the situation, to the dependent variable as the independent
variable approaches infinity. Use graphs and words to support your explanation.
For the angle of the bend (independent variable) to go to infinity means
that the inner loop is spun around and around. It will break before
turning a very, very large angle. So, as the angle of the bend goes to
infinity, the number of bends it takes to break the paperclip will
approach 0. This is evident in the graph of the function.
80
Activity 5.3 #1: Rational Functions
Activity Design:
Activity Overview:
Individual or Group
Define and locate horizontal and vertical asymptotes, interpret the meaning of
asymptotes in real-world asymptotes. Find the domain of rational functions and use
the language of rate of change to describe the behavior of rational functions.
Advance Preparation: Paper clips
Perform a stress test on each paperclip by selecting an angle at which you can repeatedly bend the paper clip
until it breaks. Repeat the stress test on each of the paper clips, using a different angle for each one.
N
a
Number of Bends
Before Breaking
Angle
45
90
180
270
360
1. Determine if the function N (a) has a constant or variable rate of change over the angles provided in the
table.
a
Angle
N
Number of Bends
Before Breaking
Rate of change
45
90
180
270
360
2. Use the language of rate of change to describe the behavior or the function on the interval from 0 to 360
degrees.
As the angle of bend increases the number of bends it takes to break decreases at a decreasing rate.
In other words, when the bend angle is small it takes a lot more bends but as the bend angle gets
large the number of bends it takes to break decreases slowly.
81
3. In the context of the situation, explain what should happen to the dependent variable as the independent
variable approaches 0.
As the independent variable (bend angle) approaches 0, the dependent variable (number of bends it
takes to break) increases dramatically.
4. Explain why it would make sense that there would be a vertical asymptote at a 0 .
When the bend angle is 0 degrees, there would be no number of bends to cause it to break.
5. In the context of the situation, explain what should happen to the dependent variable as the independent
variable approaches infinity.
As the independent variable (bend angle) approaches infinity, the dependent variable (number of
bends it takes to break) approaches 0.
6. Explain why it would make sense that there would be a horizontal asymptote at N
0.
As the angle gets larger and larger, the number of bends it takes to break approaches 0 (not even one
complete bend).
7. If the paper clip broke after 30 bends, use the table to estimate at what angle the paper clip was being bent.
Using the table above, it appears that the angle of bend would be between 45 and 90 degrees but probably
close to 50 degrees.
8. What is the practical domain for the rational function model?
The practical domain would be (0, 360) because it would not make sense to have a negative bend angle
and going much more than 360 degrees is unnecessary.
9. What is the practical range for the rational function model?
The practical range would be N greater than 0 because it would not make sense to have a negative
number of bends.
82
Activity 5.3 #2: Rational Functions
Individual or Group
Define and locate horizontal and vertical asymptotes, interpret the real-world
meaning of asymptotes, and use the language of rate of change to describe the
behavior of rational functions.
Advance Preparation: None
Activity Design:
Activity Overview:
A carnival game uses a laser, attached to a wire at a fixed height. The laser is always aimed at a mirror located on the
ground. The mirror cannot move. The object of the game is to move the laser along the wire so that the laser beam
reflects off of the mirror and hits a target which can be moved to different heights along the opposite wall. Each time
the target is hit, a prize is won and the target is moved to a different location on the wall.
D
H
In this problem context, you will investigate the relationship between the laser’s distance from the wall, D ,
and H ,the height above the ground that the laser beam will hit.
To explore the relationship between D and H , the following data were collected.
Distance from
50
75
100
125
150
175
200
wall to laser
( D in cm)
Distance from
floor to
120
65
47
38
32
28
26
reflection
( H in cm)
225
250
24
22
1. Using the data provided, create a scatterplot of the function H ( D) and draw in the function you think
would be the best fit.
2. Using the language of rate of change, give a written description of the relationship between D and H .
Answers will vary. Look for key words such as “the distance from floor decreases quickly at first as
the distance from the wall increases. This rate of decrease decreases in magnitude as the distance
from the wall to the laser continues to increase. For larger distances (say, greater than 250 cm), the
distance from floor to reflection does not change much at all for equivalent changes in the distance
from wall to laser. For smaller distances (near 25 cm), the rate of decrease is great. In fact, we see a
vertical asymptote at D = 25 cm.
83
3. The rational function model H=
( D)
a
+ c may be used to describe the relationship between D and H .
D−b
a. Using the context of the situation, determine the value of b . Explain how you know.
Since we see a vertical asymptote at D = 25 , we know that b = 25 . This is due to the
placement of the mirror 25 cm away from the wall.
b. Using the context of the situation, determine the value of c . Explain how you know.
Since we see a horizontal asymptote at H = 20 , we know that c = 20 . This is due to the fact that as
the distance of the laser from the mirror increases, the height of the reflection above the floor gets
closer and closer to 20 cm.
c. Use the data point (100, 47 ) to find the value of a .
a
+ 20
100 − 25
a
27 =
75
a = 75 ⋅ 27 = 2025
=
47
d. Write the function
H ( D) .
=
H ( D)
2025
+ 20
D − 25
4. Suppose the target is located 50 centimeters above the floor. How far from the wall shall the laser be
placed in order to hit the target?
2025
+ 20
D − 25
2025
30 =
D − 25
2025
D − 25 =
30
67.5
D − 25 =
50
=
D = 92.5 cm
5. Suppose that the target is placed 10 centimeters above the floor. Provide a mathematical argument
explaining why it would be impossible to hit the target.
Answers will vary. Students may use the graph and discuss the horizontal asymptote. Students may
make a table of values showing that the reflection height can never be 10 cm. Students may use
equations as follows:
2025
+ 20
D − 25
2025
−10 =
D − 25
2025
D − 25 =
−10
−202.5 + 25 =
−177.5
D=
=
10
The negative result is not included in our domain.
84
Activity 6.1 #1: Percentage Change
Individual or Group
Distinguish between differences (first and second) and percentage change. Calculate
growth/decay factors and percentage rates of growth.
Advance Preparation: None
Activity Design:
Activity Overview:
The following table shows the average hourly earnings of U.S. employees from 2000 through 2005. (Source:
Modeled from Statistical Abstract of the United States: 2007, Table 616).
Earnings
t
E
Time (years
since 2000)
Earnings per hour
0
1
2
3
4
5
$14.00
$14.35
$14.95
$15.35
$15.67
$16.11
1. Explain the trend of the data, using the notion of rate of change.
As time increases, the hourly earnings of employees is increasing.
2. Find the yearly growth factor and yearly percentage change in the hourly earnings from each year. Round
your answers to two decimal places.
t
E
Time (years
since 2000)
Earnings per hour
0
$14.00
1
$14.35
2
$14.95
3
$15.35
4
$15.67
5
$16.11
Growth Factor
Percentage Change
The percentage change is approximately 3% and growth factor of 1.03.
85
3. From 2000 through 2005, the inflation rate was 2.55% per year. (Source: U.S. Department of Labor,
www.bls.gov/cpi/#overview) Did hourly earnings keep pace with inflation?
Yes, earnings rose slightly faster at about 3% per year while inflation was 2.55%.
4. Estimate the average hourly earnings for 2009 using the percentage change.
For 2006 the hourly rate would be 16.11 + 0.03(16.11)
=
For 2007 the hourly rate would be 16.59 + 0.03(16.59)
=
For 2008 the hourly rate would be 17.09 + 0.03(17.09)
=
For 2009 the hourly rate would be 17.60 + 0.03(17.60)
=
16.59 or 16.11(1.03)
=
17.09 or 16.59(1.03)
=
17.60 or 17.09(1.03)
=
18.13 or 17.60(1.03)
=
$16.59 .
$17.09 .
$17.60 .
$18.13 .
5. Generate a formula for E (t ) , the earnings per hour as a function of years since 2000.
For any particular year the hourly rate would be E (t ) = 14.00(1.03)t .
6. Suppose that the hourly earnings is $17.00, estimate what year it would be using any method you choose.
We found in problem #4, for 2006 the hourly rate was
16.11 + 0.03(16.11)
= 16.59 or 16.11(1.03)
= 16.59 and for 2007 the hourly rate was
16.59 + 0.03(16.59)
= 17.09 or 16.59(1.03)
= 17.09 . Therefore, in 2007 the hourly earnings is about
$17.00.
Note that this solution does not include the use of logarithms because students have not learned how
to solve using that method as of section 6.1
86
Activity 6.1 #2: Percentage Change
Individual or Group
Distinguish between differences (first and second) and percentage change. Calculate
growth/decay factors and percentage rates of growth.
Advance Preparation: None
Activity Design:
Activity Overview:
The table of data from 2000 − 2004 below is modeled from the percentage of highway accidents in the
United States annually resulting in injuries where t is the number of years since 2000. (Source: Modeled
from Statistical Abstract of the United States 2007, Table 1047.)
1. Use a table to estimate the percentage decrease each year from 2000 to 2004.
Percent of Accidents
Years
Resulting in Injuries
(since 2000)
(percent)
t
p
0
49.85
1
47.82
2
46.49
3
45.62
4
45.06
Change Factor
Percentage Change
2. Explain the trend of the data, using the notion of rate of change.
As the years increase, the percentage of highway accidents ending in injury is decreasing at about
3% per year.
3. Estimate the percent of accidents resulting in injuries for 2007 if the trend continues.
For 2005 the percent of highway accidents ending in injury would be 45.06(0.97) = 43.71 .
For 2006 the percent of highway accidents ending in injury would be 43.71(0.97) = 42.40 .
For 2007 the percent of highway accidents ending in injury would be 42.40(0.97) = 41.13 .
87
4. Generate a formula for p (t ) , the percent of accidents resulting in injury as a function of the years since
2000.
For any particular year, the percent of highway accidents ending in injury would be
p (t ) = 49.85(0.97)t .
5. Using the data provided, sketch a graph of the function p (t ) .
6. Suppose that the value of the number of highway accidents ending in injury is 46 percent. Use the graph of
p (t ) to estimate what year it would be.
From the graph of p (t ) , the year would be 3 or 2003.
7. If the percent of highway accidents ending in injury were to decrease by 0.5 percent per year, fill in the
table below to estimate the percentage of highway accidents ending in injury from 2005 to 2008.
Years
(since 2000)
t
Percent of Accidents
Resulting in Injuries
(percent)
p
4
5
6
7
8
45.06
88
Activity 6.2 #1: Exponential Function Modeling and Graphs
Individual or Group
Construct exponential models algebraically from tables of data, predict unknown
results and interpret findings in a real-world context.
Advance Preparation: Scrap paper, Optional: Video of Mythbusters television show, “The Seven Fold
Myth”.
Activity Design:
Activity Overview:
A typical 8 ½” x 11” sheet of paper is 0.003” thick. Consider continually folding the paper in half.
1. Use a table to demonstrate the number of sheets of paper in the stack for each fold up to 7 folds.
Number of
folds
f
Number of Sheets
p
0
1
2
3
4
5
6
7
2. Estimate the thickness, in inches, of the folded paper after each fold. Do this up to 7 folds.
Number of
folds
f
Number of Sheets
n
Height of folded
paper
h
(inches)
0
1
2
3
4
5
6
7
3. Describe how the height of the folded paper is changing as the number of folds increases.
As the number of folds increases by 1, the height of the folded paper doubles.
89
4. Estimate the height of the folded paper in inches after 10 folds.
The height would continue to double from 0.384 inches.
For 8 folds, 0.384(2) = 0.768
For 9 folds, 0.768(2) = 1.536
For 10 folds, 1.536(2) = 3.072
Therefore, after 10 folds the paper should be 3.072 inches high.
5. Give a formula for the function h( f ) , the height of the paper as function of the number of folds.
We generate the height of the folded paper by taking the original height of 0.003 and doubling that
after each fold. Therefore, h( f ) = 0.003(2) f .
6. If the paper were able to be folded 50 times, how high would the paper reach?
We could answer this question by continuing the table we generated in #2 or use the function
h( f ) = 0.003(2) f . Using the function we get
h(50) = 0.003(2)50
= 3,377, 699, 721, 000 inches
7. Convert your answer to miles and compare it to the approximate distance from the Earth to the Sun of
93,000,000 miles (Source: www.newton.dep.anl.gov). Would the folded paper reach the Sun?
1 ft
1 mile
⋅
=
53,309, 654.68
12 inches 5280 feet
The folded paper would not reach the Sun but would be over halfway there!
3,377, 699, 721, 000 inches ⋅
8. Use a graph of h( f ) to approximate how many folds it would take for the folded paper to be 1,450 feet
(17,400 inches) high, which is the height of the Sears Tower in Chicago.
(Source: www.thesearstower.com)
It would only take about 22 folds to reach the height of the Sears Tower.
9. View episode 72 from the Mythbusters television show in January 2007, “The Seven Paper Fold Myth” to
see a related scientific experiment.
(Source: http://dsc.discovery.com/fansites/mythbusters/episode/episode_02.html )
90
Activity 6.2 #2: Exponential Function Modeling and Graphs
Individual or Group
Construct exponential models algebraically from tables of data Predict unknown
results and interpret findings in a real-world context.
Advance Preparation: None
Activity Design:
Activity Overview:
Your precocious paper boy offers you a new payment plan for paying him to deliver your paper. He requests
that you pay him a measly penny on the first day that he delivers your paper, two pennies on the second day,
four pennies on the third day, and so on. You are sure that there is a catch to this plan and tell him that you
will get back to him after you have had time to think about it.
1. Use a table to demonstrate the amount of money (in dollars) you would owe the paper boy for each day of
the first week.
Number of
Delivery Days
d
Paper Boy Pay
p
(in dollars)
1
2
3
4
5
6
7
2. Describe how the paper boy’s pay is changing as the number of days increases.
As the number of days increases by 1, his pay doubles.
3. Estimate the paper boy’s pay on delivery day 10.
The pay would continue to double from 64 cents on delivery day 7.
Day 8, 0.64(2) = 1.28
Day 9, 1.28(2) = 2.56
Day 10, 2.56(2) = 5.12
Therefore, after 10 delivery days the paper boy would make $5.12.
4. Give a formula for the function p (d ) , the paper boy’s pay as a function of the number of delivery days.
We generate the paper boy’s pay by doubling after each day. Therefore, p (d ) = 0.01(2) d −1 or if we
consider the initial value we can model the pay by p (d ) = 0.005(2) d .
91
5. How much would the paper boy’s salary be after 30 delivery days?
We could answer this question by continuing the table we generated in #1 or use the function
p (d ) = 0.01(2) d −1 . Using the function we get
p (30) = 0.01(2)30−1
= $5,368, 709.12
6. You realize what an outlandish amount of money that would be to pay a paperboy so you consider paying
him $0.005 (a half penny) on day one and then double his pay each day that he delivers papers thereafter.
With this adjustment, how much would the paper boy receive on delivery day 30?
p (30) = 0.005(2)30−1
.
= $2, 684,354.56
7. Use a graph of p (d ) to approximate how many days would it take him to make $50.00.
It would take about 13 days for the paper boy to make $50.00.
92
Activity 6.3 #1: Compound Interest and Continuous Growth
Activity Design:
Activity Overview:
Individual or Group
A variety of investigations of how investments can grow or decline due to compound
interest.
Advance Preparation: None
1. Suppose that your great grandfather put $100 under his mattress during the Great Depression in 1933. If
you were to find this money in 2011, it would be worth just $100. To keep up with inflation and to have
the same spending power now as it did in 1933, the money would have had to earn 3.5% interest
compounded annually. If the $100 had been invested, how much would this investment be worth in 2011?
In other words, how much money would be required in 2011 to have the same spending power as the
original $100 had in 1933?
A( y )  A0 (1  r ) y
 100(1  0.035)78
For the year 2011,
 $1463.35
2. Assume that you deposit $4500 in a bank account that pays a nominal interest rate of 4% and compounds
interest monthly. How much money will you have after 5 years? Compare this amount to the amount
you’d have if interest were paid only once per year.
Monthly:
0.04 125
A  4500(1 
)
12
 4500(1.0033)60
Yearly:
A  4500(1  0.04)5
 4500(1.04)5
 $5474.94
 $5494.48
Compounded monthly would yield $5494.48, in comparison to compounding yearly which yields $5474.94.
3. If you deposit $678 in an account with a nominal interest rate of 7.5% and continuous compounding, how
much will you have after 10 years?
A  678(e)0.07510
 678(e).75
 $1435.33
You will have $1435.33 after 10 years.
93
4. Suppose that you would like to save money for your child’s college education and could make an initial
deposit at the time of his birth in an investment with an interest rate of APR = 4.8%, compounded
annually, and leave it there for the next 18 years. How much would you have to deposit now at his birth in
order to have $50,000 after 18 years?
A
(1  APR) y
50, 000

(1  0.048)18
 $21,501.41
P
Depositing $21,501.41 now will yield the desired $50,000 in 18 years assuming that the 4.8% doesn’t
change and that you make no withdrawals or additional deposits.
5. If we assume the same set of circumstances as was described in #4 but with an APR of 7.2% and
compounding monthly how would the results be different?
A
(1  APR) y
50, 000

0.072 1218
(1 
)
12
 $13, 734.29
P
With an interest rate of 7.2%, compounding monthly, you would need to deposit $13,734.29 in order to
accumulate $50,000 after 18 years.
For problems 6 and 7, decide whether each of the statements makes sense (or is clearly true) or does not
make sense (or is clearly false).
6. Bank A was offering me an interest rate 4.7%, which is clearly a better deal for me than the 4.5% that
Bank B offered.
Not necessarily true because it also depends on the number of times compounding is done per year.
7. The bank that compounds the interest the most often will be the best place for me to deposit my money.
Not necessarily true because it also depends on the interest rate the bank offers.
94
Activity 6.3 #2: Compound Interest and Continuous Growth
Activity Design:
Activity Overview:
Individual or Group
A variety of investigations of how investments can grow or decline due to compound
interest.
Advance Preparation: None
1. Suppose that you put $50 in your piggy bank in 2010. If you had invested it in a bank account paying
3.9% interest compounded yearly, how much would it be worth in 50 years if you left the $50 untouched.
A( y )  A0 (1  r ) y
 50(1  0.039)50
 $338.65
2. Assume that you deposit $6780 in a bank account that pays a nominal interest rate of 3.6% and
compounds interest monthly. How much money will you have after 9 years? Compare this amount to the
amount you’d have if interest were paid only once per year.
Monthly:
Yearly:
0.036 129
A  6780(1 
)
12
 6780(1.003)108
A  6780(1  0.036)9
 6780(1.036)9
 $9321.11
 $9369.80
With an APR of 3.6% compounded monthly, there would be $9369.80 in the bank after 9 years,
compared to $9321.11 which is compounded yearly.
3. If you deposit $3162 in an account with a nominal interest rate of 6.3% and continuous compounding, how
much will you have after 12 years?
A  3162(e)0.06312
 3162(e).756
 $6734.24
You will have $6734.24 after 12 years, compounded continuously at an APR of 6.3%.
95
4. Suppose that you would like to save money for your child’s college education and could make an initial
deposit at the time of his birth in an investment with an interest rate of APR = 3.3%, compounded
annually, and leave it there for the next 18 years. How much would you have to deposit now at his birth in
order to have $60,000 after 18 years?
A
(1  APR) y
60, 000

(1  0.033)18
 $33, 446.10
P
Depositing $33,446.10 now will yield the desired $60,000 in 18 years, assuming that the 3.3%
doesn’t change and that you make no withdrawals or additional deposits.
5. If we assume the same set of circumstances as was described in #4 but with an APR of 6.8% and
compounding monthly, how would the results be different?
A
(1  APR) y
60, 000

0.068 1218
(1 
)
12
 $17, 704.16
P
With an interest rate of 6.8% compounded monthly, you would need to deposit $17,704.16.
For problems 6 and 7, decide whether each of the statements makes sense (or is clearly true) or does not
make sense (or is clearly false).
6. No bank could afford to pay interest every quadrillionth of a second because due to compounding, the
bank would owe an infinite number of dollars.
Not true because as the number of compoundings continues to increase the amount of dollars
reaches a limiting value. This is what we see in continuous compounding.
7. If you deposit $1000 in an investment account today the amount in the account will double to $2000 in
just over 10 years with an interest rate of 7% and compounded annually.
This is true because it will take 10 years to increase the $1000 to $1967.15 so in the 11th year the
initial investment will have doubled.
96
Activity 6.4 #1: Solving Exponential and Logarithmic Equations
Individual or Group
Solve exponential equations using logarithms and interpret the real-world meaning of
the results.
Advance Preparation: None
Activity Design:
Activity Overview:
Two of the biggest home improvement centers in the United States are Home Depot and Lowe’s. Both stores
are experiencing rapid growth, in terms of the number of stores each chain has across the country, which can
be modeled using exponential functions. Consider the following data (Source: www.homedepot.com and
www.lowes.com).
1995 1996 1997 1998 1999 2000 2001 2002 2003 2004
YEAR
Number of Home Depot Stores 423 512 624 761 930 1123 1319 1532 1707 1890
365 402 446 484 576 650 714 828 950 1087
Number of Lowe’s Stores
1. Assuming that the number of stores for each company is growing exponentially, find the average common
ratio (growth factor) for each store. What does this number tell you about the number of Home Depot and
Lowe’s Stores?
Students may approach this problem two different ways. They may compute the ratio between
successive data values and take an average of these ratios, or they may use the regression feature of
their calculator to find the exponential regression model. Either way, the intent of the question is for
students to identify the growth factor and explain what this value means.
For Home Depot, the average common ratio (growth factor) comes out 1.18 when computed by hand
(successive ratios) and 1.19 from the regression model. This number tells us that the number of Home
Depot stores is increasing by 18% or 19% each year.
For Lowe’s, the average common ratio (growth factor) comes out 1.13 when computed by hand
(successive ratios) and 1.13 from the regression model. This number tells us that the number of
Lowe’s stores is increasing by 13% each year.
2. Using your results from #1, write an exponential equation that models the data for each store. Let t = 0 be
the year 1995.
If students computed the average common ratio (growth factor) by hand (successive ratios), they
would get the following equations (models) for the number of stores. They should indicate how they
dealt with the variable T . For these results, T = 0 corresponds to 1995.
Home Depot: H (=
T ) 423 ⋅ (1.18)T
Lowe’s:
) 365 ⋅ (1.13)T
L(T
=
If students used the regression feature of their graphing calculator, the following equations (models)
would be found (if T = 0 corresponds to 1995).
Home Depot: H (=
T ) 446 ⋅ (1.19)T
Lowe’s:
L(T=
) 351 ⋅ (1.13)T
97
3. According to your models, will the number of Lowe’s Stores ever catch up to the number of Home Depot
stores? Justify your answer using complete sentences.
Since the rate of growth for Home Depot (18% per year) is greater than that of Lowe’s (13% per
year) and since Home Depot has more stores initially (in 1995) than Lowe’s, Lowe’s will never catch
up, according to the mathematical models. This can be seen graphically as well.
4. Home Depot has said that they will slow down the building of new stores and would be satisfied to have
2000 stores nationwide. According to the model, found earlier, when will the number of Home Depot
stores reach 2000? Write and solve an equation that answers this question. Show all your work, including
the use of logarithms in the solution process. Write a concluding sentence describing the results of your
work.
Using the model computed by hand we get
 2000 
T
ln 
 = ln (1.18 )
T
 423 
2000
= 423 ⋅ (1.18 )
 2000 
2000
ln 

= 1.18T
 423  ≈ 9.4
423
=
T
ln (1.18 )
Since the end of 1995 corresponds to T = 0 here, the result T = 9.4 tells us that in approximately
2005, Home Depot will have 2000 stores.
5. Lowe’s hopes to match the growth of Home Depot. Assuming that the growth continues as described by
your model, when will the number of Lowe’s stores reach 2000? Set up and solve, using any method you
like (symbolically, graphically, or numerically), the equation that would answer this question. Write a
concluding sentence describing the results of your work.
Using the model computed by hand we get
 2000 
T
ln 
 = ln (1.13 )
T
 365 
2000
= 365 ⋅ (1.13)
 2000 
2000
ln 
= 1.13T

365 

365
=
T
≈ 13.9
ln (1.13)
In about 2009 (about 14 years after 1995), Lowe’s will also have 2000 stores. According to the
problem situation, Lowe’s will catch up to Home Depot about 4 years after Home Depot reaches
2000 stores.
98
Activity 6.4 #2: Solving Exponential and Logarithmic Equations
Individual or Group
Solve exponential equations using logarithms and interpret the real-world meaning of
the results.
Advance Preparation: None
Activity Design:
Activity Overview:
The Town of Gilbert, Arizona, since 2002, has seen increases in revenue generated from its recycling
program. With an increase in population, Gilbert has also seen an increase in the amount of recyclables
collected. Furthermore, a recent newspaper article indicated that Gilbert has gone from getting $5 per ton to
$42 per ton for recyclables. Recent data showing the amount of revenue generated and the amount of
recyclables collected are shown in the table below.
Year
(Year 0 is 2000)
2
3
Revenue
$11,307
$35,583
(dollars)
Recyclables
11,477
13,285
Collected
(tons)
Source: Town of Gilbert - www.ci.gilbert.az.us
4
5
6
$48,315
$114,495
$301,633
14,460
16,088
16,644
1. Determine the exponential function that models the data representing Revenue as a function of the Year.
Use this model to predict the revenue in the year 2007.
Using the regression capability of the graphing calculator, we can find the exponential regression
t
model:=
R f=
(t ) 2639.15 ( 2.168 ) . Some instructors teach students to create this model without the
use of the regression capabilities of the graphing calculator. If students use the average ratio and use
it find the Revenue in year 0 (2000) by dividing by this common ratio, they could use the model:
t
=
R f=
(t ) 2000.35 ( 2.378 ) .
2. A major milestone for the Town of Gilbert and their recycling program would be when they generate
$1,000,000 in revenue. Use the model found in #1 above to determine when the Town of Gilbert will
generate $1,000,000 in revenue. Write and solve an equation that answers this question. Show all your
work, including the use of logarithms, in the solution process. Write a concluding sentence describing the
results of your work.
Using the model generated from the use of regression we get
 1, 000, 000 
ln 
 = t ln ( 2.168 )
 2639.15 
t
R = 2639.15 ( 2.168 )
 1, 000, 000 
ln 

t
 2639.15  = t
1, 000, 000 = 2639.15 ( 2.168 )
ln ( 2.168 )
t
 1, 000, 000 
ln 
t ≈ 7.67
 = ln ( 2.168 )
 2639.15 
According to the model, in 2008 (or earlier in 2007 according to the other model), the Town of
Gilbert will be generating $1,000,000 in revenue from its recycling program.
99
3. Determine a logarithmic function that models the data representing Recyclables Collected (in tons) as a
function of the Year. Use this function to predict the amount of recyclables that may be collected in the
year 2007.
Using the regression capabilities of the graphing calculator, the logarithmic model is:
=
T 8043.85 + 4823.46 ln x
The model may be used to predict the amount of recyclables that may be collected in the year 2007.
=
T 8043.85 + 4823.46 ln(7)
T ≈ 17430
In 2007, it is predicted that about 17,430 tons of recyclables will be collected.
4. Use your result from #1 and #3 to determine the rate in dollars per ton that the city is able to receive for its
recyclables in 2007. How does this compare with the information provided in the opening paragraph?
Explain fully.
2639.15(2.168)7
594126.63
=
≈ $34.09 per ton
8043.85 + 4823.461(ln 7) 17429.87
The newspaper reported $42 per ton. Our calculations show a rate of $34.09 per ton. The difference
may be attributed to the models we used.
5. Determine the year in which the amount of recyclables will be 20,000 tons. Using the logarithmic model
from #3, write and solve an equation that answers this question. Solve the equation using algebraic
methods. If you can solve the equation using other methods (for example, graphical), do this as well. If
you can only solve using a graphical method, for example, do so for partial credit. Write a concluding
sentence describing the results of your work.
=
20000 8043.85 + 4823.46 ln x
11956.15 = 4823.46 ln x
11956.15
= ln x
4823.46
11956.15
e 4823.46 = x
x ≈ 11.93
By 2012, the model predicts that the amount of recyclables will reach 20,000 tons.
100
Activity 6.5 #1: Logarithmic Function Modeling
Individual or Group
Use logarithmic regression to model real-world data sets and to predict and interpret
unknown results.
Advance Preparation: None
The table gives the per capita bottled water consumption for specific years (Source: U.S. Census Bureau;
Statistical Abstract of the United States, 2001; Table 204, p 130).
Activity Design:
Activity Overview:
w
Per capita Bottled
Water Consumption
(gallons)
2.4
4.5
8.0
10.7
11.6
12.5
13.1
16.0
18.1
T
Years
since 1980
0
5
10
14
15
16
17
18
19
1. Find a logarithmic function model, T = f ( w) , for the years since 1980 as a function of the per capita
bottled water consumption. (Use the first and last points in the data set to find the equation of the model
algebraically.)
 w
 w
T = log b  
T = log b  
a
a
 18.1 
 2.4 
19 = log b 
0 = log b 


 2.4 
 a 
18.1
2.4
 w 
b19 =
b0 =
=
T f=
( w) log1.11 

2.4
a
 2.4 
1
2.4
 18.1  19
1=
b=

a
 2.4 
a = 2.4
b = 1.11
2. Verify the accuracy of your results by graphing a scatter plot of the data together with the model. (Hint:
You will need to change the logarithmic function to either base e or base 10 to graph.)
101
3. Use the model to predict the year when the per capita bottled water consumption reaches 20 gallons.
 w 
=
T f=
( w) log1.11 

 2.4 
 20 
=
T f=
(20) log1.11 

 2.4 
 20 
ln 

2.4 

=
ln1.11
≈ 20.3
About 20.3 years after 1980 (in 2001), the model predicts that the per capita bottled water
consumption reached 20 gallons.
4. Use the regression feature of your calculator to find a logarithmic function model for the bottled water per
capita consumption as a function of the years since 1980. Graph the regression function along with your
function from #1.
T=
f ( w) =
−9.25 + 9.84 ln( w)
5. Use the regression model to predict the year when the per capita bottled water consumption reaches 20
gallons. Compare the results of this model with the work done in #3.
T=
f ( w) =
−9.25 + 9.84 ln( w)
T=
f (20) =
−9.25 + 9.84 ln(20)
≈ 20.23
About 20.2 years after 1980 (in 2001), the model predicts that the per capita bottled water
consumption reached 20 gallons. This is very, very close to our work done previously.
6. Determine the inverse function, w = f −1 (T ) using the function you found in #1.
 w 
T f=
=
( w) log1.11 

 2.4 
w
1.11T =
2.4
w
=
−1
(T ) 2.4 (1.11)
f=
T
w = 2.4 (1.11)
7. Use the inverse function to predict the per capita bottled water consumption in 2010.
30
−1
=
w f=
(30) 2.4 (1.11)
According to the model, the per capita bottled
water consumption will be 54.94 gallons in 2010.
w ≈ 54.94
T
102
Activity 6.5 #2: Logarithmic Function Modeling
Individual or Group
Use logarithmic regression to model real−world data sets and to predict and interpret
unknown results.
Advance Preparation: None
The data show the atmospheric pressure (in pounds per square inch) for selected altitude (in meters above sea
level) (Source: Digital Dutch 1976 Standard Atmosphere Calculator).
Activity Design:
Activity Overview:
Altitude
(meters)
m
Atmospheric pressure
(pounds per square inch)
p
1000
2000
3000
4000
5000
6000
7000
8000
13.035
11.530
10.168
8.940
7.835
6.843
5.955
5.163
1. Find a logarithmic function model, p = f (m) , for the atmospheric pressure as a function of altitude. (Use
the first and last points in the data set to find the equation of the model algebraically.)
8000 1000
=
b5.163 b13.035
m
m
p = log b  
p = log b  
b13.035 1000
a
a
=
b5.163 8000
 8000 
 1000 
5.163 = log b 
13.035 = log b 


1
 a 
 a 
b 7.872 =
8
8000
1000
b5.163 =
b13.035 =
1
a
a
 1  7.872
b= 
8000
1000
8
a = 5.163
a = 13.035
b
b
b ≈ 0.768
1000
m


Substituting=
yields a
≈ 31211.63 so then
=
p f=
(m) log 0.768 
.
13.035
0.768
 31211.63 
2. Verify the accuracy of your results by graphing a scatter plot of the data together with the model. (Hint:
You will need to change the logarithmic function to either base e or base 10 to graph.)
103
3. Use the model to predict the atmospheric pressure for Colorado Springs, CO which is at an altitude of
about 1880 meters.
m


=
p f=
(m) log 0.768 

 31211.63 
 1880 
=
p f=
(1880) log 0.768 

 31211.63 
 1880 
ln 

31211.63 
= 
ln 0.768
≈ 10.64
At an altitude of 1880 meters, the atmospheric pressure in Colorado Spring, CO is about 10.64
pounds per square inch.
4. Use the regression feature of your calculator to find a logarithmic function model for the atmospheric
pressure as a function of the altitude. Graph the regression function along with your function from #1.
=
p f=
(m) 40.39 − 3.85ln(m)
5. Use the regression model to predict the atmospheric pressure for Colorado Springs, CO which is at an
altitude of about 1880 meters. Compare the results of this model with the work done in #3.
=
p f=
(m) 40.39 − 3.85ln(m)
=
p f (1880)
= 40.39 − 3.85ln(1880)
p ≈ 11.36
At an altitude of 1880 meters, the atmospheric pressure in Colorado Spring, CO is about 11.36
pounds per square inch. This is very, very close to our work done previously.
6. Determine the inverse function, m = f −1 ( p ) using the function you found in #1.
m


f=
(m) log 0.768 

 31211.63 
m
−1
m f=
( p ) 31211.63 ⋅ 0.768 p
=
0.768 p =
31211.63
=
m 31211.63 ⋅ 0.768 p
7. Use the inverse function to predict the altitude at which the atmospheric pressure is 2 lbs. per sq. in.
−1
m f=
(2) 31211.63 ⋅ 0.7682 According to the model, at an altitude of just over 18,409
=
meters, the atmospheric pressure would be 2 lbs. per sq.
w ≈ 18409.4
in.
=
p
104
Activity 7.1 #1: Combinations of Functions
Individual or Group
Compute the sum, difference, product, or quotient of functions to model a
real-world situation and use combinations of function models to make predictions in a
real-world situation.
Advance Preparation: None
Activity Design:
Activity Overview:
The graph below shows the current trend in the amount of money spent by Americans on electricity and gas
for home use (Source: Statistical Abstract of the United States, 2008, Table 655). The graph of the function
E (t ) gives the expenditures, in billions of dollars, for electricity. The function G (t ) gives the expenditures, in
billions of dollars, for gas.
1. Explain what information E (5) provides and estimate E (5) .
E (5) tells us the expenditure for electricity by Americans in 2005. E (5) ≈ $130 billion .
2. Explain what information G (5) provides and estimate G (5) .
G (5) tells us the expenditure for electricity by Americans in 2005. G (5) ≈ $65 billion .
3. Interpret the meaning of the function ( E + G )(t ) .
The sum of the functions ( E + G )(t ) gives us the total expenditures for both electricity and gas in the
United States t years after 2000.
4. Estimate ( E + G )(5) and explain what this value means in the context of this situation. Show how the
graph is used to estimate this sum of functions.
( E + G )(5) ≈ $195 billion
In 2005, Americans spent a total of $195 billion on electricity and gas.
105
5. Sketch a graph of ( E + G )(t ) . Describe what information you can draw from this graph. That is, what
trend do you notice and what does it mean in the context of the situation?
The function ( E + G )(t ) gives the total expenditure for
both electricity and gas. The graph shows an increasing
trend. Further, the total expenditure is increasing at an
ever increasing rate.
These data suggest that possibly the cost of electricity
and gas is increasing and thus, total expenditures are
increasing. Or, Americans are using more electricity and
gas and thus, total expenditures are increasing. Or, it
could be some of both. These data don’t tell us why
( E + G )(t ) is increasing at an increasing rate.
6. Interpret the meaning of the function ( E − G )(t ) .
The function ( E − G )(t ) would tell us information about the difference between the expenditure on
electricity and the expenditure on gas. That is, it would tell us how much more Americans spend on
electricity than they spend on gas for any year after 2000.
7. Sketch a graph of ( E − G )(t ) . Describe what information you can draw from this graph. That is, what
trend do you notice and what does it mean in the context of the situation?
The function ( E − G )(t ) gives the difference in
expenditures between electricity and gas. The graph
shows an increasing trend. Further, the difference is
increasing at an ever increasing rate. However, this rate
is quite slow.
These data suggest that Americans will always spend
more on electricity than on gas. Further, this gap will
increase slowly over time.
106
Activity 7.1 #2: Combinations of Functions
Individual or Group
Compute the sum, difference, product, or quotient of functions to model a
real-world situation and use combinations of function models to make predictions in a
real-world situation.
Advance Preparation: None
Activity Design:
Activity Overview:
The function V (t )= 0.2162t 4 − 4.502t 3 + 27.63t 2 − 28.57t + 653.4 models the number of new vehicles sold
per dealership in the United States, where t is measured in years since 1995 (modeled from data from
www.autoexcemag.com). The function
=
P (t ) 777.27t + 21, 014 models the average retail selling price, P ,
of a new vehicle t years after 1995 (modeled from data from www.autoexcemag.com).
1. Explain what information V (15) provides and evaluate V (15) .
V (15) gives the number of new vehicles sold per dealership in the United States in 2010.
V (15) ≈ 2192 vehicles sold .
2. Explain what information P(15) provides and evaluate P(15) .
P (15) gives the average retail selling price of a new vehicle in 2010.
P(15) ≈ $32, 673 .
3. Evaluate V (15) ⋅ P(15) and explain what this value means in the context of the situation.
$
=
V (15) ⋅ P (15) (2192 vehicles)(32673
=
) $71, 619,325.60
vehicle
The models predict that in 2010, the total income from new car sales per car dealership will be
$71,619,325.60.
4. In general, interpret the meaning of the function V (t ) ⋅ P(t ) .
The function V (t ) ⋅ P(t ) tells the total income from new car sales per car dealership for any given
year (where t is the number of years after 1995).
107
5. Symbolically express the function V (t ) ⋅ P(t ) . Graph this function. Describe what information you can
draw from this graph. That is, what trend do you notice and what does it mean in the context of the
situation?
f=
(t ) V (t ) ⋅ P(t )
( 0.2162t − 4.502t + 27.63t − 28.57t + 653.4 ) ⋅ ( 777.27t + 21, 014 )
= ( 0.2162t ) ( 777.27t ) + ( 0.2162t ) ( 21, 014 ) + ( −4.502t ) ( 777.27t ) + ( −4.502t ) ( 21, 014 )
+ ( 27.63t ) ( 777.27t ) + ( 27.63t ) ( 21, 014 ) + ( −28.57t )( 777.27t ) + ( −28.57t )( 21, 014 )
4
=
3
2
4
2
4
3
3
2
+ ( 653.4 )( 777.27t ) + ( 653.4 )( 21, 014 )
=168.05t 5 + 4543.23t 4 − 3499.27t 4 − 94605.03t 3 + 21475.97t 3 + 580616.82t 2 − 22206.60t 2 − 600369.98t
+ 507868.22t + 13730547.60
=168.05t 5 + 1043.96t 4 − 73129.06t 3 + 558410.22t 2 − 92501.76t + 13730548.60
The total income from new car sales per car dealership is
increasing over time. This total income increases at
different rates. We might focus on what is happening as t
increases beyond t = 10 (2005). In this case, total
revenue is increasing at an ever increasing rate.
6. What is the theoretical domain and range of this function?
Domain: (−∞, ∞)
Range: (−∞, ∞)
We look at the graph of the function ignoring the context.
7. What is the practical domain and range of this function?
Answers vary.
Domain: (0,15)
Range: (14, 71.6)
Using the graph shown in #5, we choose to express the domain so that only years 1995 ( t = 0 ) to
2010 ( t = 15 ) are included. Given that domain, the range is approximately (14, 71.6) since the
minimum function value is about $14 million (when t = 0 ) and the maximum is about $71.6 million
(when t = 15 ).
108
Activity 7.2 #1: Piecewise Functions
Individual or Group
Model real-world scenarios with piecewise functions and evaluate in graphical,
formulaic, and tabular form.
Advance Preparation: None
Activity Design:
Activity Overview:
Over the course of one day’s traveling, the gallons of gasoline, g , in the tank of an automobile as a
function of time, t , in hours is displayed in the graph below.
1. After 2 hours, how many gallons of gasoline will be in the tank?
10 gallons
2. After 8 hours, how many gallons of gasoline will be in the tank?
2 gallons
3. Over what interval of hours are there 16 gallons of gasoline in the tank?
From hour 3.2 to hour 4.5.
4. Describe verbally how the amount of gasoline changed over the interval of time from 0 to 8 hours.
Over the first 3 hours, the amount of gasoline decreased at a constant rate. From 3 to 3.2
hours, gasoline was added to the tank at a constant rate. From hour3 2 to 4.5, the amount of
gasoline remained constant. From hour 4.5 to 8 the gallons of gasoline decreased at a constant
rate.
5. Explain what could possibly have occurred over the time interval from hour 3.2 to hour 4.5 on the
trip.
The automobile was either parked at a rest area, had a flat tire, or the driver stopped to have lunch.
109
6. Using the graph of the gallons of gasoline, fill in the following table.
Time
(in hours)
t
Gasoline
(in gallons)
g
0
1
2
3
4
5
6
7
8
7. Show how the graph of the function g (t ) would change if the 16 gallon tank were to be filled over
the time interval from 8 to 8.1 hours.
8. Write a formula for the piecewise function, g (t ) , over the time interval 0 to 8 hours.
 16
− 3 t +16

g (t ) = 45t − 128
16

−4t + 34
0≤t ≤3
3 < t ≤ 3.2
3.2 < t ≤ 4.5
4.5 < t ≤ 8
110
Activity 7.2 #2: Piecewise Functions
Individual or Group
Model real-world scenarios with piecewise functions and evaluate in graphical,
formulaic, and tabular form.
Advance Preparation: None
Activity Design:
Activity Overview:
The Base Disney Ticket includes admission to all four Walt Disney World Theme Parks (Magic Kingdom,
Epcot Center, Disney-MGM Studios, and Disney’s Animal Kingdom) for one park per day but does not
include admission to the Disney water parks (Pleasure Island, Disney Quest, or Wide World of Sports
Complex). Gate prices for a 5 day visit are displayed in the table below. (Source: www.wdwinfo.com)
Age
a
Gate Price
G
0<a<3
3 ≤ a < 10
10 ≤ a
Free
$190.64
$228.98
1. How much does a Disney Base ticket cost for a 12-year-old?
$228.98
2. How much does a Disney Base ticket cost for a 3-year-old?
Free
3. How much does a Disney Base ticket cost for a 7-year-old?
$190.64
4. What age group of children can buy a ticket for $190.64?
3 ≤ a < 10
Any child 3 or older, but under 10 years of age.
111
5. Describe verbally how the ticket prices changes over the different age intervals of customers.
For children up to 2-years-old are free, children 3-years-old to 9-years-old pay $190.64, and
children and adults 10-years-old and older pay $228.98 for a 5-day Base Disney Ticket.
6. A child who turns 10 tomorrow would pay how much for admission today?
$190.64
7. A child whose 10th birthday is today would pay how much for admission?
$228.98
8. Graph the function G (a ) .
9. Write a formula for the piecewise function, G (a ) .
0

=
G (a ) 190.64
228.98

112
0<t <3
3 ≤ t < 10
10 ≤ t
Activity 7.3 #1: Compositions of Functions
Individual or Group
Compose two or more functions given as a graph, formula, or table. Evaluate
composed functions and determine function composition values.
Advance Preparation: None
Activity Design:
Activity Overview:
Among other variables, automobile insurance premiums depend upon the value of the automobile, which
depends upon the age (year) of the car. Imagine that you are shopping for a used Toyota Corolla LE Sedan
with approximately 30,000 miles and need to determine how much the insurance premium will be for that
particular car. The following tables give you some important information regarding your decision. The first
table gives the 2008 clean retail value of the Corolla for different years with 30,000 miles. (Source:
www.kbb.com/KBB/UsedCars/)
Model Year 2008 Clean Retail Value of
of Vehicle
Vehicle
v
t
(with 30,000 miles)
2003
$13,140
2004
$14,240
2005
$14,935
2006
$15,080
2007
$14,205
2008
$13,760
The following table gives estimated insurance premiums for the Toyota Corolla for a specific individual
from the Allstate car insurance company (Source: www.allstate.com).
Automobile Value
v
Insurance Premium
(every 6 months)
p
$10,000 - $20,000
$20,001 - $25,000
$25,001 - $30,000
$30,001 - $40,000
$40,001 - $50,000
$470
$620
$980
$1,280
$1,405
1. Use quadratic regression to create a quadratic model to represent the 2008 value, v , of the car as a
function of the model year, t . Let t = 0 correspond to the year 2003.
v(t ) =
−250.09t 2 + 1340.16t + 13168.75
113
2. Create a graph of the insurance premium paid every six months, p ,as a function of the value of the
automobile, v . Label your axes and scale the axes appropriately.
3. It is possible to describe the situation of determining the insurance premium paid every six months, p ,
given the model year of the vehicle, t , as a composition of two functions. Explain how this situation is an
example of the composition of two functions.
The insurance premium is a function of the automobile value, p (v) , and automobile value is a
function of the model year, v(t ) . Therefore, the insurance premium is a function of the model year,
p (v(t )) .
4. What is the 2008 insurance premium charged for a 2003 Toyota Corolla LE Sedan? Express your results
using function notation.
p (v(0)) = 470
The insurance premium is $470.
5. Determine which vehicle a person could buy (by indicating the model year) if he thought he could afford
insurance premiums of $980 every 6-months. Express your results using function notation.
He could afford any vehicle because they are all in the range of $10,000 to $20,000 and the insurance
coverages are all under $980.
p (v(0)) = 470
p (v(1)) = 470
p (v(2)) = 470
.
p (v(3)) = 470
p (v(4)) = 470
p (v(5)) = 470
6. Using the formula for v(t ) found in #1, the 2008 value of a used Toyota Corolla as a function of the model
year, predict how much the insurance premiums would be for a 2005 Corolla.
p (v(2)) = 470
The premiums would be $470.
114
Activity 7.3 #2: Compositions of Functions
Group
Compose two or more functions given as a table. Interpret function compositions in a
real-world setting.
Advance Preparation: None
Activity Design:
Activity Overview:
The table below defines some functions to represent certain quantities for persons, places or things in a
school.
Function
Notation
A( x)
B( x)
C ( x)
D( x)
E ( x)
F ( x)
G ( x)
H ( x)
J ( x)
K ( x)
L( x)
Meaning
Example
Given the name of a teacher, this function
tells you the average number of students
in his/her classes.
Given the name of a school, this function
tells you the number of classrooms in the
school.
Given the name of a school, this function
tells you the number of teachers in the
school.
Given the name of a teacher, this function
tells you at which school the teacher
works.
Given the name of a teacher, this function
tells you how many years the teacher has
worked at the school.
Given a number of students, this function
tells you the number of pencils they have.
Given the total number of students a
teacher has in all of his/her classes, this
function tells you how many hours a
teacher will spend grading per week.
Given the hours a teacher spends grading
per week, this function tells you the
amount of leisure time a teacher has.
Given the number of classrooms in a
school, this function tells you the amount
of money per month the school spends on
air conditioning and/or heat.
Given the number of teachers at a school,
this function tells you the number of
students in the school.
Given the name of a teacher, this function
tells you the total number of students in
his/her classes.
115
A(Ms. Adams) = 25 students
B(Shelbyville HS) = 100 classrooms
C(Pana HS) = 65 teachers
D(Ms. Jones) = Mesquite HS
E(Mr. Smith) = 5 years
F(10 students) = 25 pencils
G(100 students) = 10 hours
H(5 hours) = 4 hours
J(50 classrooms) = $4,000
K(133 teachers) = 2500 students
L(Ms. Olivera) = 138 students
1. Use the school expressions table to create as many compositions as possible. Put these in the table below.
Your group will then be asked to present one to the class that no other group has previously presented.
Points will be given to a group each time that they are able to present a composition that has not been
previously presented.
Composition
Input
Output/Input
Output
2. Write two functions that cannot be composed and explain why.
Answers vary. One example: F ( A( x)) because even though the output to function A( x) is students it
is the AVERAGE number of students in a teacher’s class however the input to F ( x) has to be the
total number of students a teacher has in all of his/her classes. In other words, the output to the inner
function does not match up the required input of the outer function.
3. Explain the meaning of one of your compositions in the context of the real-world. Make sure to include
the input and output.
The composition G ( L( x)) means that when you input the name of the teacher the output is the
number of hours the teacher will spend grading per week.
4. Write as many “triple” compositions as possible and explain their meaning.
One answer is H (G ( L( x))) means that when you input the name of the teacher the output is the
amount of leisure time a teacher has.
116
Activity 7.4 #1: Logistic Functions
Individual or Group
Use logistic models to represent real-world data sets and to predict and interpret
unknown results. Use the language of rate of change to describe the behavior of a
logistic function.
Advance Preparation: None
Activity Design:
Activity Overview:
Based on data from 1950 to 2005, the energy consumption of the United States may be modeled by
102.388
quadrillion BTUs where t is the number of years since 1950. (Source: Modeled from
E (t ) =
1 + 3.794e −0.062t
Energy Information Administration, International Energy Annual, www.eia.doe.gov )
1. Graph E (t ) and label the independent and dependent axes. Include the domain [0,70].
2. Explain in general terms how the amount of energy consumption by the United States changes over time.
Starting in 1950 the amount of energy consumption increases at an increasing rate and then the
amount of energy consumed increases at a decreasing rate.
3. Using the graph of E (t ) , estimate the year in which the amount of energy consumption first surpassed 90
quadrillion BTUs.
From the graph of E (t ) , it appears that the amount of energy consumption first surpassed 90
quadrillion BTUs in 2000.
117
4. Does it appear that there is a limiting value for the amount of energy that the U.S. uses? If so, estimate
what this number will be. Do you think this would be reasonable?
Yes, it appears that the function E (t ) begins to approach a limiting value. It is approximately 105
quadrillion BTUs. This does not seem reasonable as the more the country grows the more likely it is
that more energy will be needed.
5. From the graph of E (t ) determine during which period from [0,70] the data is increasing and concave up.
Explain what this means about the increase in the amount of energy consumed.
It appears that the data is increasing and concave up from (0, 25). This means that the amount of
energy consumption is increasing at an increasing rate from 1950 to 1975.
6. Using the graph of E (t ) determine during which period the data is increasing and concave down. Explain
what this means about the increase in the amount of energy consumed.
It appears that the data is increasing and concave down from (25, 70). This means that the amount of
energy consumption is increasing at a decreasing rate from 1975 to 2020.
7. Using the graph estimate when the point of inflection occurs. What historical events may have been
occurring at this time to affect the amount of energy being consumed?
The point of inflection is at about (25, 58) as this is where the graph changes from concave up to
concave down. In the late 1970’s is when there were long gasoline lines and energy conservation
became important. Congress also passed a law to lower the speed limit to 55 mph on most highways.
118
Activity 7.4 #2: Logistic Functions
Individual or Group
Use logistic models to represent real-world data sets and to predict and interpret
unknown results. Use the language of rate of change to describe the behavior of a
logistic function.
Advance Preparation: None
Activity Design:
Activity Overview:
Consider a study whose goal is to model the response to a drug, r , as a function of the dose of the drug
administered in milligrams, d . The dependent variable, response, is given a value of 1 if the patient is
successfully treated by the drug and 0 if the treatment is not successful at all.
1
The function r (d ) =
models the response based upon the dosage in milligrams of a new test
1 + 545.28e −0.66 d
drug. (Source: British Pharmacological Society, www.pubmedcentral.nih.gov)
1. Graph r (d ) and label the independent and dependent axes.
2. Using the graph of r (d ) , explain how the response to the drug changes over time.
If a very small amount of the drug is administered, no patients will respond. Doubling the dose to a
larger but still small amount will not yet yield any positive response. But as the dosage increases a
threshold will be reached where the drug begins to become effective. Incremental increases in the
dosage above the threshold usually will elicit an increasingly positive effect. However, eventually a
saturation level is reached, and beyond that point increasing the dosage does not increase the
response.
3. Using the graph of r (d ) , estimate the dosage in which the response first surpasses 0.3.
From the graph of r (d ) , it appears that the response first surpasses 0.3 at 8 mg.
119
4. Does it appear that there is a limiting value for the response to the drug? If so, estimate what this number
will be.
Yes, it appears that the function r (d ) begins to approach a limiting value of r= 1 meaning a
completely successful response occurs. The limiting value occurs when a dosage of between
approximately 17 and 18 mg is given.
5. Using the graph of r (d ) determine during which period the data is increasing and concave up. Explain
what this means about the increase in the response to the drug.
It appears that the data is increasing and concave up from (0, 9). This means that the response to the
drug is increasing at an increasing rate from 0 to 9 mg.
6. Using the graph of r (d ) determine during which period the data is increasing and concave down. Explain
what this means about the increase in the response to the drug.
It appears that the data is increasing and concave down from (9, 17). This means that the response to
the drug is increasing at a decreasing rate from 9 to 17 mg.
7. Using the graph estimate when the point of inflection occurs. What might this mean in terms of the dosage
of the drug?
The point of inflection is at about (9, 0.4) as this is where the graph changes from concave up to
concave down. This would mean that before 9 mg the response was increasing at an increasing rate
and then after 9 mg of drug is administered then the response continues to increase but at a slower
rate.
120
Activity 7.5 #1: Choosing a Mathematical Model
Individual or Group
Select the “best” function to model a real-world data set and use the appropriate
function to predict and interpret unknown results.
Advance Preparation: None
Activity Design:
Activity Overview:
As a salary consultant for the Mathematics Department at a local university, you are hired to come up with a
plan for the following scenario: two professors with the same years experience, educational background, and
productivity were hired at different times and therefore a different base pay was in place when each was
hired. There is no set salary schedule in place at your university and you are asked to “catch up” the lower
paid and earlier hired professor over the next four years. How would you recommend doing so without
freezing the recently hired professor’s salary but still being fair to the professor hired earlier by getting him
to an equal salary?
Here are the particulars:
Jonathan Wickman, Ph.D., 12 years experience, hired in 2000 at the university, will receive a salary of
$84,500 for the upcoming Fall semester.
Samantha Vasquez, Ph.D., 12 years experience, hired for the upcoming Fall semester at a salary $98,750 to
draw her away from a rival university.
Show all mathematical calculations, state all assumptions, and provide detailed explanations for the decisions
you made.
---------------------------------------------------------------------------------------------------------------------Answers may vary. One possible solution follows:
Since there is no set salary schedule in place we have the freedom to set the salaries as we see fit in
the future even though the two professors will not have the same salary in the upcoming Fall
semester. We are to “catch up” Dr. Wickman up to Dr. Vasquez over a period of four years without
freezing Dr. Vasquez’ salary.
If we let t = 0 be the current year, we can use the current consumer price index which is 3.9% in
2008 (Source: U.S. Department of Labor: Bureau of Labor Statistics, www.bls.gov/CPI/) as our
assumption of how the cost of living will rise over the next four years. This will give us the
exponential mathematical model S (t ) = 98, 750(1.039)t , where S is Samantha’s salary as a function
of the number of years since the current year, t . We look at this function as a table of values and a
graph over the interval of years from t = 0 to t = 4 .
Year (since
current year)
Dr. Vasquez’ Salary
(in dollars)
t
S
0
1
2
3
4
98,750
102,601
106,603
110,760
115,080
121
Assuming a cost of living increase in her salary of 3.9% each year, Dr. Vasquez will receive $115,080 in
four years. Therefore we must work from that salary backward to get Dr. Wickman the same salary in four
years. He currently makes $84,500. If we give him a constant raise each year to “catch up” we need to
find the rate of change.
$115, 080 − $84,500 $30,580 $7, 645
= =
4−0
4 years 1 year
$7, 645
and an initial salary of $84,500 we can use a linear function to
With a constant rate of change of
1 year
model Dr. Wickman’s salary over the four years. The function=
J (t ) 84,500 + 7645t , where J is
Jonathan’s salary as a function of the number of years since the current year, t . We look at the functions,
J (t ) and S (t ) as a table of values and a graph over the interval of years from t = 0 to t = 4 . We can see
that after four years both professors will make $115, 080 therefore catching Dr. Wickman up to Dr.
Vasquez.
Year (since Dr. Wickman’s Salary Dr. Vasquez’ Salary
(in dollars)
(in dollars)
current year)
t
J
S
0
1
2
3
4
84,500
92,145
99,790
107,434
115,080
98,750
102,601
106,603
110,760
115,080
122
Activity 7.5 #2: Choosing a Mathematical Model
Individual or Group
Select the “best” function to model a real-world data set and use the appropriate
function to predict and interpret unknown results.
Advance Preparation: None
Activity Design:
Activity Overview:
The average amount of money spent by consumers on their Christmas shopping each year since 2002 is
displayed in the table below. (Source: National Retail Federation www.nrf.com )
Year
(since 2002)
t
Average Consumer
Spending on Christmas
Shopping
(in dollars)
C
0
1
2
3
4
5
649
672
702
738
791
822
1. Determine what type of mathematical model may best fit the data. Choose from linear, quadratic, or
exponential. Explain why you think the function you chose is the best choice. Be sure to justify your
choice by showing differences and ratios, a look at the scatterplot, and what would seem most reasonable
for the future. Defend your choice and refute the other two possibilities.
All three functions – linear, quadratic, and exponential – fit the data well over the interval of years
from t = 0 to t = 5 in terms of the scatterplot. In fact, the three functions are almost
indistinguishable.
If we consider differences and ratios, we see that the exponential function may fit “best” because the
ratios between consecutive years are most constant. It would also seem reasonable to use an
exponential model as most financial reports are given in percentage change.
123
2. Use regression to find the function that best fits the data and name the function C (t ) .
Choosing an exponential model for the best fit, we get the function C (t ) = 642.39(1.05)t .
3. Evaluate, give the unit of measure, and interpret C (7) .
C (7) = 642.39(1.05)7
C (7) = 903.91
In 2009, the estimated average amount of money spent on Christmas shopping is $903.91.
4. Solve and interpret the real world meaning of C ( y ) = 500 .
500 = 642.39(1.05)t
500
= 1.05t
642.39
0.778 = 1.05t
log1.05 (0.778) = t
log(0.778)
≈ −5.15
log(1.05)
Therefore, the estimated year when $500 was the average amount that consumers spent on Christmas
shopping was 6 years before 2002 or 1996.
124
Activity 8.1 #1: Using Matrices to Solve Linear Systems
Activity Design:
Individual or Group
Activity Overview: Use matrices to solve systems of equations.
Advance Preparation: None
1. The admission fee at a town’s fair is $2.00 for children and $5.00 for adults. If 2457 people
enter the fair and $7410 is collected, how many children and how many adults attended? Set up
a system of equations and use matrices to solve the system.
125
2. A landscaper places two orders with a nursery. In his first order he requests 17 bushes and 6
trees that cost a total of $885.15. In a subsequent order the landscaper purchases 8 bushes and 3
trees at a cost of $427.95. Since the bills do not list the per item price help the landscaper
determine the cost of one bush and one tree. Set up a system of equations and use matrices to
solve the system.
126
Activity 8.1 #2: Using Matrices to Solve Linear Systems
Activity Design:
Individual or Group
Activity Overview: Use matrices to solve systems of equations.
Advance Preparation: None
1. A passenger jet took four hours to fly 2008 miles with the direction of the wind and the return
flight against the wind took 4.5 hours. What was the jet’s average speed in still air and the
average speed of the wind? Set up a system of equations and use matrices to solve the system.
127
2. A week’s supply of food and medication must be purchased for the dogs and cats at a animal
shelter. Each dog’s food and medication costs 1.5 times as much as each cat’s cost. There are
currently 142 cats and 37 dogs at the shelter and budget of the shelter for such costs is $4350.
How much can be spent on each dog and each cat for food and medication?
128
Activity 8.2 #1: Matrix Operations and Applications
Activity Design:
Individual or Group
Activity Overview: Use matrix addition and scalar multiplication in real-world situations.
Advance Preparation: None
1. Derek and Reena are discussing getting married but they have outstanding loans and debts to
consider.
Derek has the following loans and debts:
$7,500 student loan to ExperiBank
$5,000 student loan to Community Bank
$9,550 car loan to State Bank
$950 credit card balance to State Bank
$275 credit card balance to Community Bank
Reena has the following loans and debts:
$8,200 student loan to ExperiBank
$6,200 student loan to Community Bank
$295 credit card balance to Community Bank
$1,100 credit card balance to State Bank
Use matrix addition to determine their combined debts of each type at the various financial
institutions.
129
2. The table below shows the base salaries of certified teachers in Tempe Union High School
District in Tempe, AZ for the 2008-2009 school year. (Source: www.tuhsd.k12.az.us )
Years Experience
BA
MA/BA+36
MA+30
MA+60
0-2
36,426
38,612
40,928
43,384
3-5
39,413
41,778
44,284
46,941
6-8
42,645
45,203
47,916
50,791
9-11
46,142
48,910
51,845
54,955
12-14
46,142
52,921
56,096
59,462
15-17
46,142
52,921
60,696
64,338
18+
46,142
52,921
60,696
69,613
Assume that for the 2009 – 2010 school year the faculty received a 1% cost of living
adjustment. Create a new salary schedule for the 2009 – 2010 school year for the Tempe
certified teachers using scalar multiplication of matrices.
130
Activity 8.2 #2: Matrix Operations and Applications
Activity Design:
Individual or Group
Activity Overview: Use matrix addition and scalar multiplication in real-world situations.
Advance Preparation: None
1. Donovan and Ruby are discussing getting married but they have outstanding loans and debts to
consider.
Donovan has the following loans and debts:
$4,500 student loan to First Bank
$6,150 student loan to Bank of Idaho
$4,650 car loan to Macon County Bank
$1875 credit card balance to Macon County Bank
$1275 credit card balance to Bank of Idaho
Ruby has the following loans and debts:
$9,100 student loan to First Bank
$3,300 student loan to Bank of Idaho
$2050 credit card balance to Bank of Idaho
$1,300 credit card balance to Macon County Bank
Use matrix addition to determine their combined debts of each type at the various financial
institutions.
131
2. Since 2002 Major League Baseball has allowed the Colorado Rockies to store their game-day
baseballs in a humidor to decrease the distance baseballs travel after being batted. The humidor
affects baseballs by decreasing the distance by 3 feet for every 10 percent increase in humidity.
The following table displays distances of batted homerun balls without being placed in the
humidor. (Source: www.sportsmemo.com )
Batter #1
362
394
426
461
401
345
351
Batter #2
386
417
452
489
329
429
325
Batter #3
409
442
479
317
360
406
401
Batter #4
333
369
407
449
402
403
396
Assume that after the baseballs are put in the humidor they travel 0.5% less. Create a table of
estimated distances for these batted balls using scalar multiplication of matrices.
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Activity 8.3 #1: Using Inverse Matrices to Solve Matrix Equations
Individual or Group
Use matrix multiplication and find inverse matrices using technology to solve
matrix equations.
Advance Preparation: None
Activity Design:
Activity Overview:
For exercises 1 – 5, perform the indicated operation. As appropriate, use the matrices
 2 5
A=

− 2 3
3 − 5
B=

 4 − 1
 4 2
C = 3 1 
5 6
1. AB
2. BA
3. CD
4. B-1
5.
A-1A
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− 1 0 4
D =  0 3 5
 2 4 0
 2 4 3
6. Find the inverse of the matrix A = 1 2 1  algebraically.
0 2 4
7. Use technology to find the inverse of the matrix A in #6 and then compare your solutions.
2 x + 2 y + 2 z = 12
1
1
3
8. Write the system of equations x − y + z =
as a matrix equation AX = B , and then
2
2
2
3
9
2x − y + z =
2
2
solve using technology.
134
Activity 8.3 #2: Using Inverse Matrices to Solve Matrix Equations
Individual or Group
Use matrix multiplication and find inverse matrices using technology to solve
matrix equations.
Advance Preparation: None
Activity Design:
Activity Overview:
For exercises 1 – 5, perform the indicated operation. As appropriate, use the matrices
 4 7
A=

− 2 1 
− 2 − 3
B=
1 
4
1
2

C=6
5 
− 1 − 3
1. AB
2. BA
3. CD
4. B-1
5.
A-1A
135
 − 2 1 2
D =  3
1 8 
 9 − 4 0
6 − 2 5 
6. Find the inverse of the matrix A = 4 2 − 1 algebraically.
0 2
3 
7. Use technology to find the inverse of the matrix A in #6 and then compare your solutions.
5x + y + 2 z = 3
8. Write the system of equations x − 2 y + 6 z = 3 as a matrix equation AX = B , and then
2 x − 2 y + 1z = 8
solve using technology.
136