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Modeling with Linear Functions

Build linear models from verbal descriptions.

Building Linear Models from Verbal Descriptions

When building linear models to solve problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
When needed, write a formula for the function.
Solve or evaluate the function using the formula.
Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
Clearly convey your result using appropriate units, and answer in full sentences when necessary.

Now let’s take a look at the student in Seattle. In Elan’s situation, there are two changing quantities: time and money. The amount of money they have remaining while on vacation depends on how long they stay. We can use this information to define our variables, including units.

\[\begin{array}{l}\text{Output:}\ M,\text{money remaining, in dollars} \\ \text{Input:}\ t,\text{time, in weeks}\end{array}\]

So, the amount of money remaining depends on the number of weeks: $M(t)$ .

\[\begin{array}{l}\text{We can also identify the initial value and the rate of change}. \\ \text{ Initial Value: She saved \$3,500, so \$3,500 is the initial value for}\ M. \\ \text{ Rate of Change: She anticipates spending \$400 each week, so}-\text{\$400 per week is the rate of change, or slope}.\end{array}\]

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

The rate of change is constant, so we can start with the linear model $M(t)=mt+b.$ Then we can substitute the intercept and slope provided.

To find the t-intercept (horizontal axis intercept), we set the output to zero, and solve for the input.

\[\begin{array}{lll}0 & = & -400t+3500 \\ t & = & \frac{3500}{400} \\ & = & 8.75\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Modeling a Set of Data with Linear Functions

Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in .

Example

Try it.

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mileRates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ . When will Keep on Trucking, Inc. be the better choice for Jamal?

Solution

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions in .

Input	$d,$ distance driven in miles
Outputs	$K(d):$ cost, in dollars, for renting from Keep on Trucking $M(d)$ cost, in dollars, for renting from Move It Your Way
Initial Value	Up-front fee: $K(0)=\text{2}0$ and $M(0)=\text{16}$
Rate of Change	$K(d)=\text{\$}0.\text{59}$ /mile and $P(d)=\text{\$}0.\text{63}$ /mile

A linear function is of the form $f(x)=mx+b.$ Using the rates of change and initial charges, we can write the equations

\[\begin{array}{lll}K(d) & = & 0.59d+20 \\ M(d) & = & 0.63d+16\end{array}\]

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when \(K(d)

These graphs are sketched in , with $K(d)$ in blue.

To find the intersection, we set the equations equal and solve:

\[\begin{array}{lll}K(d) & = & M(d) \\ 0.59d+20 & = & 0.63d+16 \\ 4 & = & 0.04d \\ \ 100 & = & d \\ d & = & 100\end{array}\]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that $K(d)$ is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is $d>100$ .

Condensed — the full section is in OpenStax College Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

The equation $F=\frac{9}{5}C+32$ is used to convert temperatures, C, on the Celsius scale to temperatures, F, on the Fahrenheit scale.
ⓐ Find the Fahrenheit temperature for a Celsius temperature of 0.
ⓑ Find the Fahrenheit temperature for a Celsius temperature of 20.
ⓒ Interpret the slope and F-intercept of the equation.
ⓓ Graph the equation.

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ⓐ
$\begin{array}{ll}\text{Find the Fahrenheit temperature for a Celsius temperature of 0.} & \ F=\frac{9}{5}C+32 \\ \text{Find}\ F\ \text{when}\ C=0. & \ F=\frac{9}{5}(0)+32 \\ \text{Simplify.} & \ F=32\end{array}$
ⓑ
$\begin{array}{ll}\text{Find the Fahrenheit temperature for a Celsius temperature of 20.} & \ F=\frac{9}{5}C+32 \\ \text{Find}\ F\ \text{when}\ C=20. & \ F=\frac{9}{5}(20)+32 \\ \text{Simplify.} & \ F=36+32 \\ \text{Simplify.} & \ F=68\end{array}$
ⓒ
Interpret the slope and F-intercept of the equation.
Even though this equation uses F and C, it is still in slope–intercept form.
The slope, $\frac{9}{5},$ means that the temperature Fahrenheit (F) increases 9 degrees when the temperature Celsius (C) increases 5 degrees.
The F-intercept means that when the temperature is $0\text{^{\circ}}$ on the Celsius scale, it is $32\text{^{\circ}}$ on the Fahrenheit scale.
ⓓ Graph the equation.
We’ll need to use a larger scale than our usual. Start at the F-intercept $(0,32)$ , and then count out the rise of 9 and the run of 5 to get a second point as shown in the graph.
Janis is planning to rent a car while on vacation. The equation C(m)=.25m+10 models the relationship between the cost in dollars, C, per day and the number of miles,m, she drives in one day.
ⓐ Find the cost if Janis drives the car 20 miles one day.
ⓑ Find the cost on a day when Janis drives the car 400 miles.
ⓒ Interpret the slope and y-intercept (or C intercept) of the equation in terms of the variables and units used in this problem.
ⓓ Graph the linear function below. Be sure to show the scale you are using on your coordinate system.
A function that will convert women’s dress sizes in the US, x, to dress sizes in Italy, I(x), is given by:
$I(x)=2(x+10)$
ⓐ
x, US women’s size $I(x)=2(x+10)$ Italian women’s size
4
10
16
ⓑ Interpret the slope and y-intercept (or I intercept) for this linear function in terms of the variables used in the problem.
ⓒ Graph the function below. Be sure to show the scale you are using.
Edwin pays a monthly fee for water service on his apartment of $18 plus an additional $0.30 for each HCF (hundred cubic feet) of water used.
ⓐ Write a cost function, C(w), which will give his monthly cost as a function of water used, w, in hundred cubic feet.
ⓑ
w (in HCF) C(w) (in $)
65
80
ⓒ On a certain month Edwin’s bill seems high at $50. How many HCF’s of water did Edwin use that month.
ⓓ Graph the function below. Be sure to show the scale you are using.
Cassandra is a botanist who is studying the growth of pea plants under a variety of conditions. One experiment yields the following results. For a seedling starting off at 3 inches the growth rate each week 1.25 inches.
ⓐWrite a height function, H(t), which will give the plant height as a function of time in weeks, t.
ⓑ Estimate the plant height for each of the following amounts of time paying attention to units provided.
t H(t)
7 days
14 days
2.5 weeks
ⓒ Her plants will need to be secured to a stick when they reach 12 inches in height. Estimate the time when her samples will need to be secured.
ⓓ Graph the function below. Be sure to show the scale you are using.
A town’s population has been growing linearly. In 2004, the population was 6,200. By 2009, the population had grown to 8,100. Assume this trend continues.
1. ⓐPredict the population in 2013.
2. ⓑIdentify the year in which the population will reach 15,000.
Kusonyeza yankho

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2000 years ago!
To make computation a little nicer, we will define our input as the number of years since 2004.
\[\begin{array}{l} \\ \begin{array}{l}\text{Input:}\ t,\text{years since 2004} \\ \text{Output:}\ P(t),\text{the town’s population}\end{array}\end{array}\]
To predict the population in 2013 ( $t=9$ ), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.
To determine the rate of change, we will use the change in output per change in input.
\[m=\frac{\text{change in output}}{\text{change in input}}\]
The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to $t=0,$ giving the point $(0,\text{6200}).$ Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to $t=\text{5,}$ giving the point $(5,\text{8100}).$
The two coordinate pairs are $(0,\text{6200})$ and $(5,\text{8100}).$ Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.
\[\begin{array}{lll}m & = & \frac{8100-6200}{5-0} \\ & = & \frac{1900}{5} \\ & = & 380\ \text{people per year}\end{array}\]
We already know the y-intercept of the line, so we can immediately write the equation:
\[P(t)=380t+6200\]
To predict the population in 2013, we evaluate our function at $t=9.$
\[\begin{array}{lll}P(9) & = & 380(9)+6,200 \\ & = & 9,620\end{array}\]
If the trend continues, our model predicts a population of 9,620 in 2013.
To find when the population will reach 15,000, we can set $P(t)=15000$ and solve for $t.$
\[\begin{array}{lll}15000 & = & 380t+6200 \\ 8800 & = & 380t \\ t & \approx & 23.158\end{array}\]
Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.
A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.
ⓐ Write a linear model to represent the cost $C$ of the company as a function of $x,$ the number of doughnuts produced.
ⓑ Find and interpret the y-intercept.

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ⓐ $C(x)=0.25x+25,000$
ⓑ The y-intercept is $(0,25,000)$. If the company does not produce a single doughnut, they still incur a cost of $25,000.
A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.
1. ⓐPredict the population in 2014.
2. ⓑIdentify the year in which the population will reach 54,000.
Kusonyeza yankho

ⓐ41,100 ⓑ2020
Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

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In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question:
"How long will it take them to be 2 miles apart"?
In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.
\[\begin{array}{l} \\ \begin{array}{l}\text{Input:}\ t,\text{time in hours}. \\ \text{Output:}\ A(t),\text{distance in miles, and}\ E(t),\text{distance in miles}\end{array}\end{array}\]
Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as .
Initial Value: They both start at the same intersection so when $t=0,$ the distance traveled by each person should also be 0. Thus the initial value for each is 0.
Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for $A$ is 4 and the slope for $E$ is 3.
Using those values, we can write formulas for the distance each person has walked.
\[\begin{array}{lll}A(t) & = & 4t \\ E(t) & = & 3t\end{array}\]
For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, $A,$ which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, $E,$ to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.
We can then define a third variable, $D,$ to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from .
Recall that we need to know how long it takes for $D,$ the distance between them, to equal 2 miles. Notice that for any given input $t,$ the outputs $A(t),E(t),$ and $D(t)$ represent distances.
shows us that we can use the Pythagorean Theorem because we have drawn a right angle.
Using the Pythagorean Theorem, we get:
\[\begin{array}{llll}D{(t)}^{2} & = & A{(t)}^{2}+E{(t)}^{2} & \\ & = & {(4t)}^{2}+{(3t)}^{2} & \\ & = & 16{t}^{2}+9{t}^{2} & \\ & = & 25{t}^{2} & \\ \ D(t) & = & \pm \sqrt{25{t}^{2}} & \ \text{Solve for }D(t)\ \text{using the square root.} \\ & = & \pm 5|t| & \end{array}\]
In this scenario we are considering only positive values of $t,$ so our distance $D(t)$ will always be positive. We can simplify this answer to $D(t)=5t.$ This means that the distance between Anna and Emanuel is also a linear function. Because $D$ is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output $D(t)=2$ and solve for $t.$
\[\begin{array}{lll}D(t) & = & 2 \\ 5t & = & 2 \\ t & = & \frac{2}{5}=0.4\end{array}\]
They will fall out of radio contact in 0.4 hour, or 24 minutes.
There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

Kusonyeza yankho

It might help here to draw a picture of the situation. See . It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates $(\text{3}0,\text{1}0),$ and Eastborough at $(\text{2}0,\ 0).$
Using this point along with the origin, we can find the slope of the line from Westborough to Agritown.
\[m=\frac{10-0}{30-0}=\frac{1}{3}\]
Now we can write an equation to describe the road from Westborough to Agritown.
\[W(x)=\frac{1}{3}x\]
From this, we can determine the perpendicular road to Eastborough will have slope $m=-3.$ Because the town of Eastborough is at the point (20, 0), we can find the equation.
\[\begin{array}{llll}E(x) & = & -3x+b & \\ 0 & = & -3(20)+b & \ \text{Substitute }(20,0)\ \text{into the equation}. \\ b & = & 60 & \\ E(x) & = & -3x+60 & \end{array}\]
We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,
\[\begin{array}{llll}\ \frac{1}{3}x & = & -3x+60 & \\ \frac{10}{3}x & = & 60 & \\ 10x & = & 180 & \\ x & = & 18 & \ \text{Substitute this back into }W(x). \\ y & = & W(18) & \\ & = & \frac{1}{3}(18) & \\ & = & 6 & \end{array}\]
The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.
\[\begin{array}{lll}\text{distance} & = & \sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}} \\ & = & \sqrt{{(18-0)}^{2}+{(6-0)}^{2}} \\ & \approx & \ 18.974\ \text{miles}\end{array}\]
There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

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21.57 miles

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The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions in .

Input	$d,$ distance driven in miles
Outputs	$K(d):$ cost, in dollars, for renting from Keep on Trucking $M(d)$ cost, in dollars, for renting from Move It Your Way
Initial Value	Up-front fee: $K(0)=\text{2}0$ and $M(0)=\text{16}$
Rate of Change	$K(d)=\text{\$}0.\text{59}$ /mile and $P(d)=\text{\$}0.\text{63}$ /mile

A linear function is of the form $f(x)=mx+b.$ Using the rates of change and initial charges, we can write the equations

\[\begin{array}{lll}K(d) & = & 0.59d+20 \\ M(d) & = & 0.63d+16\end{array}\]

These graphs are sketched in , with $K(d)$ in blue.

To find the intersection, we set the equations equal and solve:

\[\begin{array}{lll}K(d) & = & M(d) \\ 0.59d+20 & = & 0.63d+16 \\ 4 & = & 0.04d \\ \ 100 & = & d \\ d & = & 100\end{array}\]

Explain how to find the input variable in a word problem that uses a linear function.

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Determine the independent variable. This is the variable upon which the output depends.
Explain how to find the output variable in a word problem that uses a linear function.
Explain how to interpret the initial value in a word problem that uses a linear function.

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To determine the initial value, find the output when the input is equal to zero.
Explain how to determine the slope in a word problem that uses a linear function.
Find the area of a parallelogram bounded by the y-axis, the line $x=3,$ the line $f(x)=1+2x,$ and the line parallel to $f(x)$ passing through $(\text{2},\text{7}).$

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6 square units
Find the area of a triangle bounded by the x-axis, the line $f(x)=12-\frac{1}{3}x,$ and the line perpendicular to $f(x)$ that passes through the origin.
Find the area of a triangle bounded by the y-axis, the line $f(x)=9-\frac{6}{7}x,$ and the line perpendicular to $f(x)$ that passes through the origin.

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20.01 square units
Find the area of a parallelogram bounded by the x-axis, the line $g(x)=2,$ the line $f(x)=3x,$ and the line parallel to $f(x)$ passing through $(6,1).$
Predict the population in 2016.

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2,300
Identify the year in which the population will reach 0.
Predict the population in 2016.

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64,170
Identify the year in which the population will reach 75,000.
Find the linear function that models the town’s population $P$ as a function of the year, $t,$ where $t$ is the number of years since the model began.

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$P(t)=75,000+2500t$
Find a reasonable domain and range for the function $P.$
If the function $P$ is graphed, find and interpret the x- and y-intercepts.

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(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.
If the function $P$ is graphed, find and interpret the slope of the function.
When will the population reach 100,000?

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Ten years after the model began
What is the population in the year 12 years from the onset of the model?
Find the linear function that models the baby’s weight $W$ as a function of the age of the baby, in months, $t.$

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$W(t)=0.5t+7.5$
Find a reasonable domain and range for the function $W.$
If the function $W$ is graphed, find and interpret the x- and y-intercepts.

Kusonyeza yankho

$(-15,\ 0)$ : The x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. $(0,\text{7}.\text{5})$ : The baby weighed 7.5 pounds at birth.
If the function W is graphed, find and interpret the slope of the function.
When did the baby weigh 10.4 pounds?

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At age 5.8 months
What is the output when the input is 6.2?
Find the linear function that models the number of people inflicted with the common cold $C$ as a function of the year, $t.$

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$C(t)=12,025-205t$
Find a reasonable domain and range for the function $C.$
If the function $C$ is graphed, find and interpret the x- and y-intercepts.

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$(\text{58}.\text{7},\ 0):$ In roughly 59 years, the number of people inflicted with the common cold would be 0. $(0,\text{12},0\text{25})$ Initially there were 12,025 people afflicted by the common cold.
If the function $C$ is graphed, find and interpret the slope of the function.

Symbols used here

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Modeling with Linear Functions

Build linear models from verbal descriptions.
Model a set of data with a linear function.
Graph and interpret applications of slope–intercept form of a linear function. (IA 3.2.5)
Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
When needed, write a formula for the function.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Sankhani wanu

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Input	\(d,\) distance driven in miles
Outputs	\(K(d):\) cost, in dollars, for renting from Keep on Trucking \(M(d)\) cost, in dollars, for renting from Move It Your Way
Initial Value	Up-front fee: \(K(0)=\text{2}0\) and \(M(0)=\text{16}\)
Rate of Change	\(K(d)=\text{\$}0.\text{59}\) /mile and \(P(d)=\text{\$}0.\text{63}\) /mile

x, US women’s size	\(I(x)=2(x+10)\) Italian women’s size
4
10
16

w (in HCF)	C(w) (in $)
65
80

t	H(t)
7 days
14 days
2.5 weeks