Section1.4Linear Functions

Functions whose graphs are straight lines are both the simplest and the most important functions in mathematics. Lines often model important phenomena, and even when they don't directly model phenomena, lines can often approximate other functions that do. Whether a function's graph is a straight line or not is connected directly to its average rate of change.

Preview Activity1.4.1.
1. Let $y = f(x) = 7 - 3x\text{.}$ Determine $AV_{[-3,-1]}\text{,}$ $AV_{[2,5]}\text{,}$ and $AV_{[4,10]}$ for the function $f\text{.}$

2. Let $y = g(x)$ be given by the data in Table 1.4.1.

Determine $AV_{[-5,-2]}\text{,}$ $AV_{[-1,1]}\text{,}$ and $AV_{[0,4]}$ for the function $g\text{.}$

3. Consider the function $y = h(x)$ defined by the graph in Figure 1.4.2. Figure 1.4.2. The graph of $y = h(x)\text{.}$

Determine $AV_{[-5,-2]}\text{,}$ $AV_{[-1,1]}\text{,}$ and $AV_{[0,4]}$ for the function $h\text{.}$

4. What do all three examples above have in common? How do they differ?

5. For the function $y = f(x) = 7 - 3x$ from (a), find the simplest expression you can for

\begin{equation*} AV_{[a,b]} = \frac{f(b)-f(a)}{b-a} \end{equation*}

where $a \ne b\text{.}$

Subsection1.4.1Properties of linear functions

In Preview Activity 1.4.1, we considered three different functions for which the average rate of change of each appeared to always be constant. For the first function in the preview activity, $y = f(x) = 7 - 3x\text{,}$ we can compute its average rate of change on an arbitrary interval $[a,b]\text{.}$ Doing so, we notice that

\begin{align*} AV_{[a,b]} &= \frac{f(b)-f(a)}{b-a}\\ &= \frac{(7-3b) - (7-3a)}{b-a}\\ &= \frac{7-3b - 7+3a}{b-a}\\ &= \frac{-3b + 3a}{b-a}\\ &= \frac{-3(b - a)}{b-a}\\ &= -3\text{.} \end{align*}

This result shows us that for the function $y = f(x) = 7 - 3x\text{,}$ its average rate of change is always $-3\text{,}$ regardless of the interval we choose. We will use the property of having constant rate of change as the defining property of a linear function.

Definition1.4.3.

A function $f$ is linear provided that its average rate of change is constant on every choice of interval in its domain 1 . That is, for any inputs $a$ and $b$ for which $a \ne b\text{,}$ it follows that

\begin{equation*} \frac{f(b) - f(a)}{b-a} = m \end{equation*}

for some fixed constant $m\text{.}$ We call $m$ the slope of the linear function $f\text{.}$

Here we are considering functions whose domain is the set of all real numbers.

From prior study, we already know a lot about linear functions. In this section, we work to understand some familiar properties in light of the new perspective of Definition 1.4.3.

Let's suppose we know that a function $f$ is linear with average rate of change $AV_{[a,b]} = m$ and that we also know the function value is $y_0$ at some fixed input $x_0\text{.}$ That is, we know that $f(x_0) = y_0\text{.}$ From this information, we can find the formula for $y = f(x)$ for any input $x\text{.}$ Working with the known point $(x_0, f(x_0))$ and any other point $(x,f(x))$ on the function's graph, we know that the average rate of change between these two points must be the constant $m\text{.}$ This tells us that

\begin{equation*} \frac{f(x) - f(x_0)}{x-x_0} = m\text{.} \end{equation*}

Since we are interested in finding a formula for $y = f(x)\text{,}$ we solve this most recent equation for $f(x)\text{.}$ Multiplying both sides by $(x-x_0)\text{,}$ we see that

\begin{equation*} f(x) - f(x_0) = m(x-x_0)\text{.} \end{equation*}

Adding $f(x_0)$ to each side, it follows

\begin{equation} f(x) = f(x_0) + m(x-x_0)\text{.}\label{E-linear-point-slope-first-take}\tag{1.4.1} \end{equation}

This shows that to determine the formula for a linear function, all we need to know is its average rate of change (or slope) and a single point the function passes through.

Example1.4.4.

Find a formula for a linear function $f$ whose average rate of change is $m = -\frac{1}{4}$ and passes through the point $(-7,-5)\text{.}$

Solution. Using Equation (1.4.1) and the facts that $m = -\frac{1}{4}$ and $f(-7) = -5$ (that is, $x_0 = -7$ and $f(x_0) = -5$), we have

\begin{equation*} f(x) = -5 -\frac{1}{4}(x - (-7)) = 5 - \frac{1}{4}(x+7)\text{.} \end{equation*}

Replacing $f(x)$ with $y$ and $f(x_0)$ with $y_0\text{,}$ we call Equation (1.4.1) the point-slope form of a line.

Point-slope form of a line.

A line with slope $m$ (equivalently, average rate of change $m$) that passes through the point $(x_0,y_0)$ has equation

\begin{equation*} y = y_0 + m(x-x_0)\text{.} \end{equation*}
Activity1.4.2.

Find an equation for the line that is determined by the following conditions; write your answer in point-slope form wherever possible.

1. The line with slope $\frac{3}{7}$ that passes through $(-11, -17)\text{.}$

2. The line passing through the points $(-2,5)$ and $(3,-1)\text{.}$

3. The line passing through $(4,9)$ that is parallel to the line $2x - 3y = 5\text{.}$

4. Explain why the function $f$ given by Table 1.4.5 appears to be linear and find a formula for $f(x)\text{.}$ Figure 1.4.6. Plot of a linear function $h\text{.}$
5. Find a formula for the linear function shown in Figure 1.4.6.

Visualizing the various components of point-slope form is important. For a line through $(x_0,y_0)$ with slope $m\text{,}$ we know its equation is $y = y_0 + m(x-x_0)\text{.}$ In Figure 1.4.7, we see that the line passes through $(x_0,y_0)$ along with an arbitary point $(x,y)\text{,}$ which makes the vertical change between the two points given by $y - y_0$ and the horizontal change between the points $x - x_0\text{.}$ This is consistent with the fact that

\begin{equation*} AV_{[x_0,x]} = m = \frac{y-y_0}{x-x_0}\text{.} \end{equation*}

Indeed, writing $m = \frac{y-y_0}{x-x_0}$ is a rearrangement of the point-slope form of the line, $y = y_0 + m(x-x_0)\text{.}$

We naturally use the terms “increasing” and “decreasing” as from Definition 1.3.7 to describe lines based on whether their slope is positive or negative. A line with positive slope, such as the one in Figure 1.4.7, is increasing because its constant rate of change is positive, while a line with negative slope, such as in Figure 1.4.8 is decreasing because of its negative rate of change. We say that a horizontal line (one whose slope is $m = 0$) is neither increasing nor decreasing.

A special case arises when the known point on a line satisfies $x_0 = 0\text{.}$ In this situation, the known point lies on the $y$-axis, and thus we call the point the “$y$-intercept” of the line. The resulting form of the line's equation is called slope-intercept form, which is also demonstrated in Figure 1.4.8.

Slope-intercept form.

For the line with slope $m$ and passing through $(0,y_0)\text{,}$ its equation is

\begin{equation*} y = y_0 + mx\text{.} \end{equation*}

Slope-intercept form follows from point-slope form from the fact that replacing $x_0$ with $0$ gives us $y = y_0 + m(x-0) = y_0 + mx\text{.}$ In many textbooks, the slope-intercept form of a line (often written $y = mx + b$) is treated as if it is the most useful form of a line. Point-slope form is actually more important and valuable since we can easily write down the equation of a line as soon as we know its slope and any point that lies on it, as opposed to needing to find the $y$-intercept, which is needed for slope-intercept form. Moreover, point-slope form plays a prominent role in calculus.

If a line is in slope-intercept or point-slope form, it is useful to be able to quickly interpret key information about the line from the form of its equation.

Example1.4.9.

For the line given by $y = -3 - 2.5(x-5)\text{,}$ determine its slope and a point that lies on the line.

Solution. This line is in point-slope form. Its slope is $m = -2.5$ and a point on the line is $(5,-3)\text{.}$

Example1.4.10.

For the line given by $y = 6 + 0.25x\text{,}$ determine its slope and a point that lies on the line.

Solution. This line is in slope-intercept form. Its slope is $m = 0.25$ and a point on the line is $(0,6)\text{,}$ which is also the line's $y$-intercept.

Subsection1.4.2Interpreting linear functions in context

Since linear functions are defined by the property that their average rate of change is constant, linear functions perfectly model quantities that change at a constant rate. In context, we can often think of slope as a rate of change; analyzing units carefully often yields significant insight.

Example1.4.11.

The Dolbear function $T = D(N) = 40 + 0.25N$ from Section 1.2 is a linear function whose slope is $m = 0.25\text{.}$ What is the meaning of the slope in this context?

Solution

Recall that $T$ is measured in degrees Fahrenheit and $N$ in chirps per minute. We know that $m = AV_{[a,b]} = 0.25$ is the constant average rate of change of $D\text{.}$ Its units are “units of output per unit of input”, and thus “degrees Fahrenheit per chirp per minute”. This tells us that the average rate of change of the temperature function is $0.25$ degrees Fahrenheit per chirp per minute, which means that for each additional chirp per minute observed, we expect the temperature to rise by $0.25$ degrees Fahrenheit.

Indeed, we can observe this through function values. We note that $T(60) = 55$ and $T(61) = 55.25\text{:}$ one additional observed chirp per minute corresponds to a $0.25$ degree increase in temperature. We also see this in the graph of the line, as seen in Figure 1.4.12: the slope between the points $(40,50)$ and $(120,70)$ is

\begin{align*} m &= \frac{70-50}{120-40}\\ &= \frac{20}{80}\\ &= 0.25 \frac{\text{degrees F}}{\text{chirp per minute}}\text{.} \end{align*} Figure 1.4.12. The linear Dolbear function with slope $m = 0.25$ degrees Fahrenheit per chirp per minute.

Like with the Dolbear function, it is often useful to write a linear function (whose output is called $y$) that models a quantity changing at a constant rate (as a function of some input $t$) by writing the function relationship in the form

\begin{equation*} y = b + mt \end{equation*}

where $b$ and $m$ are constants. We may think of the four quantities involved in the following way:

• The constant $b$ is the “starting value” of the output that corresponds to an input of $t = 0\text{;}$

• The constant $m$ is the rate at which the output changes with respect to changes in the input: for each additional $1$-unit change in input, the output will change by $m$ units.

• The variable $t$ is the independent (input) variable. A nonzero value for $t$ corresponds to how much the input variable has changed from an initial value of $0\text{.}$

• The variable $y$ is the dependent (output) variable. The value of $y$ results from a particular choice of $t\text{,}$ and can be thought of as the starting output value ($b$) plus the change in output that results from the corresponding change in input $t\text{.}$

Activity1.4.3.

The summit of Africa's largest peak, Mt. Kilimanjaro 2 , has two main ice fields and a glacier at its peak. Geologists measured the ice cover in the year 2000 ($t = 0$) to be approximately $1951$ m$^2\text{;}$ in the year 2007, the ice cover measured $1555$ m$^2\text{.}$

The main context of this problem comes from Exercise 30 on p. 27 of Connally's Functions Modeling Change, 5th ed.
1. Suppose that the amount of ice cover at the peak of Mt. Kilimanjaro is changing at a constant average rate from year to year. Find a linear model $A = f(t)$ whose output is the area, $A\text{,}$ in square meters in year $t$ (where $t$ is the number of years after 2000).

2. What do the slope and $A$-intercept mean in the model you found in (a)? In particular, what are the units on the slope?

3. Compute $f(17)\text{.}$ What does this quantity measure? Write a complete sentence to explain.

4. If the model holds further into the future, when do we predict the ice cover will vanish?

5. In light of your work above, what is a reasonable domain to use for the model $A = f(t)\text{?}$ What is the corresponding range?

Activity1.4.4.

In each of the following prompts, we investigate linear functions in context.

1. A town's population initially has $28750$ people present and then grows at a constant rate of $825$ people per year. Find a linear model $P = f(t)$ for the number of people in the town in year $t\text{.}$

2. A different town's population $Q$ is given by the function $Q = g(t) = 42505 - 465t\text{.}$ What is the slope of this function and what is its meaning in the model? Write a complete sentence to explain.

3. A spherical tank is being drained with a pump. Initially the tank is full with $\frac{32\pi}{3}$ cubic feet of water. Assume the tank is drained at a constant rate of $1.2$ cubic feet per minute. Find a linear model $V = p(t)$ for the total amount of water in the tank at time $t\text{.}$ In addition, what is a reasonable domain for the model?

4. A conical tank is being filled in such a way that the height of the water in the tank, $h$ (in feet), at time $t$ (in minutes) is given by the function $h = q(t) = 0.65t\text{.}$ What can you say about how the water level is rising? Write at least one careful sentence to explain.

5. Suppose we know that a $5$-year old car's value is $$10200\text{,}$ and that after $10$ years its value is$$4600\text{.}$ Find a linear function $C = L(t)$ whose output is the value of the car in year $t\text{.}$ What is a reasonable domain for the model? What is the value and meaning of the slope of the line? Write at least one careful sentence to explain.

Subsection1.4.3Summary

• Any function $f$ with domain all real numbers that has a constant average rate of change on every interval $[a,b]$ will have a straight line graph. We call such functions linear functions.

• A linear function $y = f(x)$ can be written in the form $y = f(x) = y_0 + m(x-x_0)\text{,}$ where $m$ is the slope of the line and $(x_0,y_0)$ is a point that lies on the line. In particular, $f(x_0) = y_0\text{.}$

• In an applied context where we have a linear function that models a phenomenon in the world around us, the slope tells us the function's (constant) average rate of change. The units on the slope, $m\text{,}$ are always “units of output per unit of input” and this enables us to articulate how the output changes in response to a $1$-unit change in input.

Exercises1.4.4Exercises

6.

An apartment manager keeps careful record of how the rent charged per unit corresponds to the number of occupied units in a large complex. The collected data is shown in Table 1.4.13. 3

This problem is a slightly modified version of one found in Carroll College's Chapter Zero resource for Active Calculus.
1. Why is it reasonable to say that the number of occupied apartments is a linear function of rent?

2. Let $A$ be the number of occupied apartments and $R$ the monthly rent charged (in dollars). If we let $A = f(R)\text{,}$ what is the slope of the linear function $f\text{?}$ What is the meaning of the slope in the context of this question?

3. Determine a formula for $A = f(R)\text{.}$ What do you think is a reasonable domain for the function? Why?

4. If the rent were to be increased to $1000, how many occupied apartments should the apartment manager expect? How much total revenue would the manager collect in a given month when rent is set at$1000?

5. Why do you think the apartment manager is interested in the data that has been collected?

7.

Alicia and Dexter are each walking on a straight path. For a particular $10$-second window of time, each has their velocity (in feet per second) measured and recorded as a function of time. Their respective velocity functions are plotted in Figure 1.4.14. Figure 1.4.14. The velocity functions $A = f(t)$ and $D = g(t)$ for Alicia and Damon, respectively.
1. Determine formulas for both $A = f(t)$ and $D = g(t)\text{.}$

2. What is the value and meaning of the slope of $A\text{?}$ Write a complete sentence to explain and be sure to include units in your response.

3. What is the value and meaning of the average rate of change of $D$ on the interval $[4,8]\text{?}$ Write a complete sentence to explain and be sure to include units in your response.

4. Is there ever a time when Alicia and Damon are walking at the same velocity? If yes, determine both the time and velocity; if not, explain why.

5. Is is possible to determine if there is ever a time when Alicia and Damon are located at the same place on the path? If yes, determine the time and location; if not, explain why not enough information is provided.

8.

An inverted conical tank with depth $4$ feet and radius $2$ feet is completely full of water. The tank is being drained by a pump in such a way that the amount of water in the tank is decreasing at a constant rate of $1.5$ cubic feet per minute. Let $V = f(t)$ denote the volume of water in the tank at time $t$ and $h = g(t)$ the depth of the water in the tank at time $t\text{,}$ where $t$ is measured in minutes.

1. How much water is in the tank at $t = 0$ when the tank is completely full?
2. Explain why volume, $V\text{,}$ when viewed as a function of time, $t\text{,}$ is a linear function.
3. Determine a formula for $V = f(t)\text{.}$
5. What is a reasonable domain to use for the model $f\text{?}$ What is its corresponding range?