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Chapter 2:
Linear Functions
Chapter one was a window that gave us a peek into the entire course. Our goal was to
understand the basic structure of functions and function notation, the toolkit functions,
domain and range, how to recognize and understand composition and transformations of
functions and how to understand and utilize inverse functions. With these basic
components in hand we will further research the specific details and intricacies of each
type of function in our toolkit and use them to model the world around us.
Mathematical Modeling
As we approach day to day life we often need to quantify the things around us, giving structure and numeric value to various situations. This ability to add structure enables
us to make choices based on patterns we see that are weighted and systematic. With
this structure in place we can model and even predict behavior to make decisions.
Adding a numerical structure to a real world situation is called Mathematical
Modeling.
When modeling real world scenarios, there are some common growth patterns that are
regularly observed. We will devote this chapter and the rest of the book to the study of
the functions used to model these growth patterns.
Section 2.1 Linear Functions ...................................................................................... 101
Section 2.2 Graphs of Linear Functions ..................................................................... 114
Section 2.3 Modeling with Linear Functions .............................................................. 129 Section 2.4 Fitting Linear Models to Data .................................................................. 141
Section 2.5 Absolute Value Functions ........................................................................ 149
Section 2.1 Linear Functions
As you hop into a taxicab in Las Vegas, the meter will immediately read $3. 5 0; this is the
“drop” charge made when the taximeter is activated. After that initial fee, the taximeter
will add $2. 76 for each mile the taxi drives^1. In this scenario, the total taxi fare depends
upon the number of miles ridden in the taxi, and we can ask whether it is possible to
model this type of scenario with a function. Using descriptive variables, we choose m for
miles and C for Cost in dollars as a function of miles: C(m).
(^1) Nevada Taxicab Authority, retrieved Aug 4, 2020. There is also a waiting fee assessed when the taxi is
waiting at red lights, but we’ll ignore that in this discussion.
102 Chapter 2
We know for certain that C ( 0 )= 3. 50 , since the $3. 5 0 drop charge is assessed regardless
of how many miles are driven. Since $2. 67 is added for each mile driven, then
C ( 1 )= 3. 50 + 2. 67 = 6. 17.
If we then drove a second mile, another $2. 67 would be added to the cost:
C ( 2 )= 3. 50 + 2. 67 + 2. 67 = 3. 50 + 2. 67 ( 2 )= 8. 84
If we drove a third mile, another $2. 67 would be added to the cost:
C ( 3 )= 3. 50 + 2. 67 + 2. 67 + 2. 67 = 3. 50 + 2. 67 ( 3 )= 11. 51
From this we might observe the pattern, and conclude that if m miles are driven,
C ( m )= 3. 50 + 2. 67 m because we start with a $3. 5 0 drop fee and then for each mile
increase we add $2. 67.
It is good to verify that the units make sense in this equation. The $3. 5 0 drop charge is
measured in dollars; the $2. 67 charge is measured in dollars per mile.
( mmiles ) mile
dollars C m dollars
When dollars per mile are multiplied by a number of miles, the result is a number of
dollars, matching the units on the 3. 5 0, and matching the desired units for the C function.
Notice this equation C ( m )= 3. 50 + 2. 67 m consisted of two quantities. The first is the
fixed $3. 5 0 charge which does not change based on the value of the input. The second is
the $2. 67 dollars per mile value, which is a rate of change. In the equation, this rate of
change is multiplied by the input value.
Looking at this same problem in table format we can also see the cost changes by $2. 67
for every 1 mile increase.
m 0 1 2 3
C(m) 3. 50 6.17 8.84 11.
It is important here to note that in this equation, the rate of change is constant ; over any
interval, the rate of change is the same.
Graphing this equation, C ( m )= 3. 50 + 2. 67 m we see
the shape is a line, which is how these functions get
their name: linear functions.
When the number of miles is zero the cost is $3. 5 0,
giving the point (0, 3. 5 0) on the graph. This is the
vertical or C(m) intercept. The graph is increasing in a
straight line from left to right because for each mile
the cost goes up by $2. 67 ; this rate remains consistent.
104 Chapter 2
N(m) is an increasing linear function. With this formula we can predict how many
songs he will have in 1 year (12 months):
N ( 12 )= 200 + 15 ( 12 )= 200 + 180 = 380. Marcus will have 380 songs in 12 months.
Try it Now
- If you earn $30,000 per year and you spend $29,000 per year write an equation for the
amount of money you save after y years, if you start with nothing. “The most important thing, spend less than you earn!^2 ”
Calculating Rate of Change
Given two values for the input, x 1 and x 2 , and two corresponding values for the output,
y 1 and y 2 , or a set of points, ( x 1 , y 1 )and ( x 2 , y 2 ), if we wish to find a linear
function that contains both points we can calculate the rate of change, m :
2 1
2 1
changein input
changeinoutput
x x
y y
x
y m −
Rate of change of a linear function is also called the slope of the line.
Note in function notation, y 1 = f ( x 1 ) and y 2 = f ( x 2 ), so we could equivalently write
( 2 ) ( 1 )
2 1
f x f x m x x
Example 2
The population of a city increased from 23,400 to 27,800 between 2002 and 2006. Find
the rate of change of the population during this time span.
The rate of change will relate the change in population to the change in time. The
population increased by 27800 − 23400 = 4400 people over the 4 year time interval. To
find the rate of change, the number of people per year the population changed by:
year
people
years
people 1100 4
=^ = 1100 people per year
Notice that we knew the population was increasing, so we would expect our value for m
to be positive. This is a quick way to check to see if your value is reasonable.
(^2) http://www.thesimpledollar.com/2009/06/19/rule- 1 - spend-less-than-you-earn/
Section 2.1 Linear Functions 105
Example 3
The pressure, P , in pounds per square inch (PSI) on a diver depends upon their depth
below the water surface, d , in feet, following the equation P ( d )= 14. 696 + 0. 434 d.
Interpret the components of this function.
The rate of change, or slope, 0.434 would have units ft
PSI
depth
pressure
input
output = =. This
tells us the pressure on the diver increases by 0.434 PSI for each foot their depth
increases.
The initial value, 14.696, will have the same units as the output, so this tells us that at a depth of 0 feet, the pressure on the diver will be 14.696 PSI.
Example 4
If f ( x )is a linear function, f ( 3 )=− 2 , and f ( 8 )= 1 , find the rate of change.
f ( 3 )=− 2 tells us that the input 3 corresponds with the output - 2, and f ( 8 )= 1 tells us
that the input 8 corresponds with the output 1. To find the rate of change, we divide the
change in output by the change in input:
changeininput
change inoutput
−
m = =. If desired we could also write this as m = 0.
Note that it is not important which pair of values comes first in the subtractions so long
as the first output value used corresponds with the first input value used.
Try it Now
- Given the two points (2, 3) and (0, 4), find the rate of change. Is this function
increasing or decreasing?
We can now find the rate of change given two input-output pairs, and can write an
equation for a linear function once we have the rate of change and initial value. If we
have two input-output pairs and they do not include the initial value of the function, then
we will have to solve for it.
Section 2.1 Linear Functions 107
Example 7
Working as an insurance salesperson, Ilya earns a base salary and a commission on each
new policy, so Ilya’s weekly income, I , depends on the number of new policies, n , he
sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies, and earned $920. Find an equation for I(n) ,
and interpret the meaning of the components of the equation.
The given information gives us two input-output pairs: (3,760) and (5,920). We start
by finding the rate of change.
m =
Keeping track of units can help us interpret this quantity. Income increased by $
when the number of policies increased by 2, so the rate of change is $80 per policy; Ilya
earns a commission of $80 for each policy sold during the week.
We can then solve for the initial value
I ( n )= b + 80 n then when n = 3, I (3) = 760 , giving
760 = b + 80 ( 3 ) this allows us to solve for b
b = 760 − 80 ( 3 )= 520
This value is the starting value for the function. This is Ilya’s income when n = 0,
which means no new policies are sold. We can interpret this as Ilya’s base salary for
the week, which does not depend upon the number of policies sold.
Writing the final equation: I ( n )= 520 + 80 n
Our final interpretation is: Ilya’s base salary is $520 per week and he earns an
additional $80 commission for each policy sold each week.
Flashback
Looking at Example 7 :
Determine the independent and dependent variables. What is a reasonable domain and range?
Is this function one-to-one?
Try it Now
- The balance in your college payment account, C , is a function of the number of
quarters, q , you attend. Interpret the function C(a) = 20000 – 4000 q in words. How
many quarters of college can you pay for until this account is empty?
108 Chapter 2
Example 8
Given the table below write a linear equation that represents the table values
We can see from the table that the initial value of rats is 1000 so in the linear format
P w ( ) = b + mw , b = 1000.
Rather than solving for m , we can notice from the table that the population goes up by
80 for every 2 weeks that pass. This rate is consistent from week 0, to week 2, 4, and 6. The rate of change is 80 rats per 2 weeks. This can be simplified to 40 rats per week and
we can write
P w ( ) = b + mw as P ( w )= 1000 + 40 w
If you didn’t notice this from the table you could still solve for the slope using any two
points from the table. For example, using (2, 1080) and (6, 1240),
1240 1080 160 40 6 2 4
m
rats per week
Important Topics of this Section
Definition of Modeling
Definition of a linear function
Structure of a linear function
Increasing & Decreasing functions
Finding the vertical intercept (0, b )
Finding the slope/rate of change, m
Interpreting linear functions
Try it Now Answers
- S ( y )= 30 , 000 y − 29 , 000 y = 1000 y $1000 is saved each year.
m = ; Decreasing because m < 0
- Your College account starts with $20,000 in it and you withdraw $4,000 each quarter
(or your account contains $20,000 and decreases by $4000 each quarter.) Solving C(a) = 0 gives a = 5. You can pay for 5 quarters before the money in this
account is gone.
w , number of
weeks
P(w) , number
of rats
110 Chapter 2
Section 2.1 Exercises
- A town's population has been growing linearly. In 2003, the population was 45,000,
and the population has been growing by 1700 people each year. Write an equation,
P t ( ) , for the population t years after 2003.
- A town's population has been growing linearly. In 2005, the population was 69,000,
and the population has been growing by 2500 people each year. Write an equation,
P t ( ) , for the population t years after 2005.
- Sonya is currently 10 miles from home, and is walking further away at 2 miles per
hour. Write an equation for her distance from home t hours from now.
- A boat is 100 miles away from the marina, sailing directly towards it at 10 miles per
hour. Write an equation for the distance of the boat from the marina after t hours.
- Timmy goes to the fair with $40. Each ride costs $2. How much money will he have
left after riding n rides?
- At noon, a barista notices she has $20 in her tip jar. If she makes an average of $0.
from each customer, how much will she have in her tip jar if she serves n more
customers during her shift?
Determine if each function is increasing or decreasing
- f ( x ) = 4 x + 3 8. g ( x ) = 5 x + 6
- a x ( ) = 5 − 2 x 10. b x ( ) = 8 − 3 x
- h x ( (^) ) = − 2 x + 4 12. k (^) ( x (^) ) = − 4 x + 1
- ( )
j x = x − 14. ( )
p x = x −
- ( )
n x = − x − 16. ( )
m x = − x +
Find the slope of the line that passes through the two given points
- (2, 4) and (4, 10) 18. (1, 5) and (4, 11)
- (-1,4) and (5, 2) 20. (-2, 8) and (4, 6)
- (6,11) and (-4,3) 22. (9,10) and (-6,-12)
Section 2.1 Linear Functions 111
Find the slope of the lines graphed
- Sonya is walking home from a friend’s house. After 2 minutes she is 1.4 miles from
home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate?
- A gym membership with two personal training sessions costs $125, while gym
membership with 5 personal training sessions costs $260. What is the rate for
personal training sessions?
- A city's population in the year 1960 was 287,500. In 1989 the population was
275,900. Compute the slope of the population growth (or decline) and make a
statement about the population rate of change in people per year.
- A city's population in the year 1958 was 2,113,000. In 1991 the population was
2,099,800. Compute the slope of the population growth (or decline) and make a
statement about the population rate of change in people per year.
- A phone company charges for service according to the formula: C n ( )^^ =^24 +0.1 n ,
where n is the number of minutes talked, and C n ( ) is the monthly charge, in dollars.
Find and interpret the rate of change and initial value.
- A phone company charges for service according to the formula: C n ( ) = 26 +0.04 n ,
where n is the number of minutes talked, and C n ( ) is the monthly charge, in dollars.
Find and interpret the rate of change and initial value.
- Terry is skiing down a steep hill. Terry's elevation, E t ( ), in feet after t seconds is
given by E t ( ) = 3000 − 70 t. Write a complete sentence describing Terry’s starting
elevation and how it is changing over time.
Section 2.1 Linear Functions 113
- A farmer finds there is a linear relationship between the number of bean stalks, n , she
plants and the yield, y , each plant produces. When she plants 30 stalks, each plant
yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans.
Find a linear relationships in the form y = mn + b that gives the yield when n stalks
are planted.
- Which of the following tables could represent a linear function? For each that could
be linear, find a linear equation models the data.
x g(x)
0 5
5 - 10
10 - 25
15 - 40
x h(x)
0 5
5 30
10 105
15 230
x f(x)
0 - 5
5 20
10 45
15 70
x k(x)
5 13
10 28
20 58
25 73
- Which of the following tables could represent a linear function? For each that could
be linear, find a linear equation models the data.
x g(x)
0 6
2 - 19
4 - 44
6 - 69
x h(x)
2 13
4 23
8 43
10 53
x f(x)
2 - 4
4 16
6 36
8 56
x k(x)
0 6
2 31
6 106
8 231
- While speaking on the phone to a friend in Oslo, Norway, you learned that the current
temperature there was - 23 Celsius (- 23
o C). After the phone conversation, you wanted
to convert this temperature to Fahrenheit degrees, oF, but you could not find a
reference with the correct formulas. You then remembered that the relationship
between oF and oC is linear. [UW]
a. Using this and the knowledge that 32
o F = 0
o C and 212
o F = 100
o C, find an
equation that computes Celsius temperature in terms of Fahrenheit; i.e. an
equation of the form C = “an expression involving only the variable F.”
b. Likewise, find an equation that computes Fahrenheit temperature in terms of
Celsius temperature; i.e. an equation of the form F = “an expression involving
only the variable C.”
c. How cold was it in Oslo in oF?
114 Chapter 2
Section 2.2 Graphs of Linear Functions
When we are working with a new function, it is useful to know as much as we can about
the function: its graph, where the function is zero, and any other special behaviors of the
function. We will begin this exploration of linear functions with a look at graphs.
When graphing a linear function, there are three basic ways to graph it:
By plotting points (at least 2) and drawing a line through the points
Using the initial value (output when x = 0) and rate of change (slope)
Using transformations of the identity function f ( x )= x
Example 1
Graph f x x 3
( )= 5 − by plotting points
In general, we evaluate the function at two or more inputs to find at least two points on
the graph. Usually it is best to pick input values that will “work nicely” in the equation.
In this equation, multiples of 3 will work nicely due to the 3
in the equation, and of
course using x = 0 to get the vertical intercept. Evaluating f(x) at x = 0 , 3 and 6:
f
f
f
These evaluations tell us that the points (0,5), (3,3), and (6,1) lie on the graph of the line.
Plotting these points and drawing a line through
them gives us the graph.
When using the initial value and rate of change to graph, we need to consider the
graphical interpretation of these values. Remember the initial value of the function is the
output when the input is zero, so in the equation f ( x )= b + mx , the graph includes the
point (0, b ). On the graph, this is the vertical intercept – the point where the graph
crosses the vertical axis.
For the rate of change, it is helpful to recall that we calculated this value as
changeofinput
changeofoutput m =
116 Chapter 2
Try it Now
- Consider that the slope
− could also be written as
. Using
, find another
point on the graph that has a negative x value.
Another option for graphing is to use transformations of the identity function f ( x )= x.
In the equation f ( x )= mx , the m is
acting as the vertical stretch of the
identity function. When m is
negative, there is also a vertical
reflection of the graph. Looking at
some examples:
In f ( x )= mx + b , the b acts as the
vertical shift, moving the graph up
and down without affecting the
slope of the line. Some examples:
Section 2.2 Graphs of Linear Functions 117
Using Vertical Stretches or Compressions along with Vertical Shifts is another way to
look at identifying different types of linear functions. Although this may not be the
easiest way for you to graph this type of function, make sure you practice each method.
Example 3
Graph f x x 2
( )=− 3 + using transformations.
The equation is the graph of the identity function vertically compressed by ½ and
vertically shifted down 3.
Vertical compression combined with Vertical shift
Notice how this nicely compares to the other method where the vertical intercept is found
at (0, - 3) and to get to another point we rise (go up vertically) by 1 unit and run (go
horizontally) by 2 units to get to the next point (2, - 2), and the next one (4, - 1). In these
three points (0, - 3), (2, - 2), and (4, - 1), the output values change by +1, and the x values
change by +2, corresponding with the slope m = ½.
Example 4
Match each equation with one of the lines in the graph below
j x x
hx x
gx x
f x x
Only one graph has a vertical intercept of -
3, so we can immediately match that graph with g(x).
Section 2.2 Graphs of Linear Functions 119
There are two special cases of lines: a horizontal line and a
vertical line. In a horizontal line like the one graphed to the
right, notice that between any two points, the change in the
outputs is 0. In the slope equation, the numerator will be 0,
resulting in a slope of 0. Using a slope of 0 in the
f ( x )= b + mx , the equation simplifies to f ( x )= b.
Notice a horizontal line has a vertical intercept, but no
horizontal intercept (unless it’s the line f(x) = 0).
In the case of a vertical line, notice that between any two
points, the change in the inputs is zero. In the slope equation,
the denominator will be zero, and you may recall that we
cannot divide by the zero; the slope of a vertical line is
undefined. You might also notice that a vertical line is not a
function. To write the equation of vertical line, we simply
write input=value, like x = b.
Notice a vertical line has a horizontal intercept, but no vertical
intercept (unless it’s the line x = 0).
Horizontal and Vertical Lines
Horizontal lines have equations of the form f ( x )= b
Vertical lines have equations of the form x = a
Example 6
Write an equation for the horizontal line graphed above.
This line would have equation f ( ) x = 2
Example 7
Write an equation for the vertical line graphed above.
This line would have equation x = 2
Try it Now
- Describe the function f ( x )= 6 − 3 x in terms of transformations of the identity
function and find its horizontal intercept.
120 Chapter 2
Parallel and Perpendicular Lines
When two lines are graphed together, the lines will be parallel if they are increasing at
the same rate – if the rates of change are the same. In this case, the graphs will never
cross (unless they’re the same line).
Parallel Lines
Two lines are parallel if the slopes are equal (or, if both lines are vertical).
In other words, given two linear equations f ( x )= b + m 1 x and g ( x )= b + m 2 x , the
lines will be parallel if m 1 = m 2.
Example 8
Find a line parallel to f ( x )= 6 + 3 x that passes through the point (3, 0)
We know the line we’re looking for will have the same slope as the given line, m = 3.
Using this and the given point, we can solve for the new line’s vertical intercept:
g ( x )= b + 3 x then at (3, 0),
b
b
The line we’re looking for is g ( x )=− 9 + 3 x
If two lines are not parallel, one other interesting possibility is that the lines are
perpendicular, which means the lines form a right angle (90 degree angle – a square
corner) where they meet. In this case, the slopes when multiplied together will equal - 1.
Solving for one slope leads us to the definition:
Perpendicular Lines
Given two linear equations f ( x )= b + m 1 x and g ( x )= b + m 2 x
The lines will be perpendicular if m 1 m 2 =− 1 , and so
1
2
m
m
We often say the slope of a perpendicular line is the “negative reciprocal” of the other
line’s slope.
