Linear Functions — Precalculus 2e

Representing Linear Functions

The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.

Representing a Linear Function in Word Form

Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.

The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.

The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.

Representing a Linear Function in Function Notation

Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the slope-intercept form of a line, where $x$ is the input value, $m$ is the rate of change, and $b$ is the initial value of the dependent variable.

\begin{array}{l} Equation form & y = m x + b \\ Function notation & f (x) = m x + b \end{array}

In the example of the train, we might use the notation $D (t)$ in which the total distance $D$ is a function of the time $t .$ The rate, $m,$ is 83 meters per second. The initial value of the dependent variable $b$ is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.

D (t) = 83 t + 250

Representing a Linear Function in Tabular Form

A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.

Table with the first row, labeled t, containing the seconds from 0 to 3, and with the second row, labeled D(t), containing the meters 250 to 499. The first row goes up by 1 second, and the second row goes up by 83 meters. — Figure 2 Tabular representation of the function D showing selected input and output values

Representing a Linear Function in Graphical Form

Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, $D (t) = 83 t + 250,$ to draw a graph, represented in Figure 3. Notice the graph is a line. When we plot a linear function, the graph is always a line.

The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph in Figure 3 that the y-intercept in the train example we just saw is $(0, 250)$ and represents the distance of the train from the station when it began moving at a constant speed.

A graph of an increasing function with points at (-2, -4) and (0, 2). — Figure 3 The graph of $D (t) = 83 t + 250.$ Graphs of linear functions are lines because the rate of change is constant.

Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line $f (x) = 2 x +1.$ Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.

Example 1

Using a Linear Function to Find the Pressure on a Diver

The pressure, $P,$ in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, $d,$ in feet. This relationship may be modeled by the equation, $P (d) = 0.434 d + 14.696.$ Restate this function in words.

Scuba diver. — Figure 4 (credit: Ilse Reijs and Jan-Noud Hutten)

Solution

To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.

Analysis

The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.

Determining whether a Linear Function Is Increasing, Decreasing, or Constant

The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 5(a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5(c).

Three graphs depicting an increasing function, a decreasing function, and a constant function. — Figure 5

Example 2

Deciding whether a Function Is Increasing, Decreasing, or Constant

Studies from the early 2010s indicated that teens sent about 60 texts a day, while more recent data indicates much higher messaging rates among all users, particularly considering the various apps with which people can communicate. For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.

ⓐ The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.
ⓑ A person has a limit of 500 texts per month in their data plan. The input is the number of days, and output is the total number of texts remaining for the month.
ⓒ A person has an unlimited number of texts in their data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.

Solution

Analyze each function.

ⓐ The function can be represented as $f (x) = 60 x$ where $x$ is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.
ⓑ The function can be represented as $f (x) = 500 - 60 x$ where $x$ is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after $x$ days.
ⓒ The cost function can be represented as $f (x) = 50$ because the number of days does not affect the total cost. The slope is 0 so the function is constant.

Calculating and Interpreting Slope

In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Given two values for the input, $x_{1}$ and $x_{2},$ and two corresponding values for the output, $y_{1}$ and $y_{2}$ —which can be represented by a set of points, $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ —we can calculate the slope $m,$ as follows

m = \frac{change in output (rise)}{change in input (run)} = \frac{Δ y}{Δ x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

where $Δ y$ is the vertical displacement and $Δ x$ is the horizontal displacement. Note in function notation two corresponding values for the output $y_{1}$ and $y_{2}$ for the function $f,$ $y_{1} = f (x_{1})$ and $y_{2} = f (x_{2}),$ so we could equivalently write

m = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}}

Figure 6 indicates how the slope of the line between the points, $(x_{1,} y_{1})$ and $(x_{2,} y_{2}),$ is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.

Graph depicting how to calculate the slope of a line — Figure 6 The slope of a function is calculated by the change in $y$ divided by the change in $x .$ It does not matter which coordinate is used as the $(x_{2,} y_{2})$ and which is the $(x_{1}, y_{1}),$ as long as each calculation is started with the elements from the same coordinate pair.

Example 3

Finding the Slope of a Linear Function

If $f (x)$ is a linear function, and $(3,−2)$ and $(8,1)$ are points on the line, find the slope. Is this function increasing or decreasing?

Solution

The coordinate pairs are $(3,−2)$ and $(8,1) .$ To find the rate of change, we divide the change in output by the change in input.

m = \frac{change in output}{change in input} = \frac{1 - (- 2)}{8 - 3} = \frac{3}{5}

We could also write the slope as $m = 0.6.$ The function is increasing because $m > 0.$

Analysis

As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used.

Example 4

Finding the Population Change from a Linear Function

The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012.

Solution

The rate of change relates the change in population to the change in time. The population increased by $27, 800 - 23, 400 = 4,400$ people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.

\frac{4,400 people}{4 years} = 1,100 \frac{people}{year}

So the population increased by 1,100 people per year.

Analysis

Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.

Writing the Point-Slope Form of a Linear Equation

Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the point-slope form.

y - y_{1} = m (x - x_{1})

The point-slope form is derived from the slope formula.

\begin{array}{l} m = \frac{y - y_{1}}{x - x_{1}} & assuming x \neq x_{1} \\ m (x - x_{1}) = \frac{y - y_{1}}{x - x_{1}} (x - x_{1}) & Multiply both sides by (x - x_{1}) . \\ m (x - x_{1}) = y - y_{1} & Simplify . \\ y - y_{1} = m (x - x_{1}) & Rearrange . \end{array}

Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, $y - 4 = - \frac{1}{2} (x - 6)$ . We can convert it to the slope-intercept form as shown.

\begin{array}{l} y - 4 = - \frac{1}{2} (x - 6) \\ y - 4 = - \frac{1}{2} x + 3 & Distribute the - \frac{1}{2} . \\ y = - \frac{1}{2} x + 7 & Add 4 to each side . \end{array}

Therefore, the same line can be described in slope-intercept form as $y = - \frac{1}{2} x + 7.$

Writing the Equation of a Line Using a Point and the Slope

The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told that a line has a slope of 2 and passes through the point $(4, 1) .$ We know that $m = 2$ and that $x_{1} = 4$ and $y_{1} = 1.$ We can substitute these values into the general point-slope equation.

\begin{matrix} y - y_{1} = m (x - x_{1}) \\ y - 1 = 2 (x - 4) \end{matrix}

If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.

\begin{array}{l} y - 1 = 2 (x - 4) \\ y - 1 = 2 x - 8 & Distribute the 2. \\ y = 2 x - 7 & Add 1 to each side . \end{array}

Both equations, $y - 1 = 2 (x - 4)$ and $y = 2 x - 7,$ describe the same line. See Figure 7.

A Cartesian coordinate system displays a straight blue line with an upward positive slope. The x-axis ranges from -8 to 8, and the y-axis ranges from -8 to 8, with grid lines and labels at integer intervals. Three distinct points are marked on the line: the first at coordinates (2, -7), the second at (3, 0) which is the x-intercept, and the third at (4, 1). The line extends indefinitely in both directions as indicated by arrows on its ends. — Figure 7

Example 5

Writing Linear Equations Using a Point and the Slope

Write the point-slope form of an equation of a line with a slope of 3 that passes through the point $(6, –1) .$ Then rewrite it in the slope-intercept form.

Solution

Let’s figure out what we know from the given information. The slope is 3, so $m = 3.$ We also know one point, so we know $x_{1} = 6$ and $y_{1} = −1.$ Now we can substitute these values into the general point-slope equation.

\begin{array}{l} y - y_{1} = m (x - x_{1}) \\ y - (- 1) = 3 (x - 6) & Substitute known values . \\ y + 1 = 3 (x - 6) & Distribute - 1 to find point-slope form . \end{array}

Then we use algebra to find the slope-intercept form.

\begin{array}{l} y + 1 = 3 (x - 6) \\ y + 1 = 3 x - 18 & Distribute 3 . \\ y = 3 x - 19 & Simplify to slope-intercept form . \end{array}

Writing the Equation of a Line Using Two Points

The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points $(0, 1)$ and $(3, 2) .$ We can use the coordinates of the two points to find the slope.

\begin{array}{l} \begin{array}{l} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{array} \\ = \frac{2 - 1}{3 - 0} \\ = \frac{1}{3} \end{array}

Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point.

\begin{matrix} y - y_{1} = m (x - x_{1}) \\ y - 1 = \frac{1}{3} (x - 0) \end{matrix}

As before, we can use algebra to rewrite the equation in the slope-intercept form.

\begin{array}{l} y - 1 = \frac{1}{3} (x - 0) \\ y - 1 = \frac{1}{3} x & Distribute the \frac{1}{3} . \\ y = \frac{1}{3} x + 1 & Add 1 to each side . \end{array}

Both equations describe the line shown in Figure 8.

A graph on a coordinate plane shows a line passing through the points (-3, 0) and (0, 1). The x and y axes range from -4 to 4. — Figure 8

Example 6

Writing Linear Equations Using Two Points

Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form.

Solution

Let’s begin by finding the slope.

\begin{array}{l} \begin{array}{l} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{array} \\ = \frac{7 - 1}{8 - 5} \\ = \frac{6}{3} \\ = 2 \end{array}

So $m = 2.$ Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use $(5, 1) .$

\begin{matrix} y - y_{1} = m (x - x_{1}) \\ y - 1 = 2 (x - 5) \end{matrix}

The point-slope equation of the line is $y_{2} - 1 = 2 (x_{2} - 5) .$ To rewrite the equation in slope-intercept form, we use algebra.

\begin{array}{l} y - 1 = 2 (x - 5) \\ y - 1 = 2 x - 10 \\ y = 2 x - 9 \end{array}

The slope-intercept equation of the line is $y = 2 x - 9.$

Writing and Interpreting an Equation for a Linear Function

Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function $f$ in Figure 9.

We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose $(0, 7)$ and $(4, 4) .$ We can use these points to calculate the slope.

\begin{array}{l} \begin{array}{l} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{array} \\ \begin{array}{l} = \frac{4 - 7}{4 - 0} \\ = - \frac{3}{4} \end{array} \end{array}

Now we can substitute the slope and the coordinates of one of the points into the point-slope form.

\begin{array}{l} y - y_{1} = m (x - x_{1}) \\ y - 4 = - \frac{3}{4} (x - 4) \end{array}

If we want to rewrite the equation in the slope-intercept form, we would find

\begin{array}{l} y - 4 = - \frac{3}{4} (x - 4) \\ y - 4 = - \frac{3}{4} x + 3 \\ y = - \frac{3}{4} x + 7 \end{array}

If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, $b = 7.$ We now have the initial value $b$ and the slope $m$ so we can substitute $m$ and $b$ into the slope-intercept form of a line.

The image demonstrates the substitution of values for the slope (m) and y-intercept (b) into the general form of a linear function, f(x) = mx + b. An arrow points from -3/4 to 'm', and another arrow points from 7 to 'b', indicating that m = -3/4 and b = 7. Below this, the specific linear function f(x) = -3/4x + 7 is shown as the result of these substitutions.

So the function is $f (x) = - \frac{3}{4} x + 7,$ and the linear equation would be $y = - \frac{3}{4} x + 7.$

Example 7

Writing an Equation for a Linear Function

Write an equation for a linear function given a graph of $f$ shown in Figure 10.

Graph of an increasing function with points at (-3, 0) and (0, 1). — Figure 10

Solution

Identify two points on the line, such as $(0, 2)$ and $(- 2, −4) .$ Use the points to calculate the slope.

\begin{array}{l} \begin{array}{l} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{array} \\ = \frac{- 4 - 2}{- 2 - 0} \\ = \frac{- 6}{- 2} \\ = 3 \end{array}

Substitute the slope and the coordinates of one of the points into the point-slope form.

\begin{matrix} y - y_{1} = m (x - x_{1}) \\ y - (- 4) = 3 (x - (- 2)) \\ y + 4 = 3 (x + 2) \end{matrix}

We can use algebra to rewrite the equation in the slope-intercept form.

\begin{array}{l} y + 4 = 3 (x + 2) \\ y + 4 = 3 x + 6 \\ y = 3 x + 2 \end{array}

Analysis

This makes sense because we can see from Figure 11 that the line crosses the y-axis at the point $(0, 2)$ , which is the y-intercept, so $b = 2.$

Graph of an increasing line with points at (0, 2) and (-2, -4). — Figure 11

Example 8

Writing an Equation for a Linear Cost Function

Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function $C$ where $C (x)$ is the cost for $x$ items produced in a given month.

Solution

The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by $37.5.$ The cost Ben incurs is the sum of these two costs, represented by $C (x) = 1250 + 37.5 x .$

Analysis

If Ben produces 100 items in a month, his monthly cost is represented by

\begin{array}{l} C (100) = 1250 + 37.5 (100) \\ = 5000 \end{array}

So his monthly cost would be $5,000.

Example 9

Writing an Equation for a Linear Function Given Two Points

If $f$ is a linear function, with $f (3) = −2$ , and $f (8) = 1$ , find an equation for the function in slope-intercept form.

Solution

We can write the given points using coordinates.

\begin{array}{l} f (3) = −2 \to (3, −2) \\ f (8) = 1 \to (8, 1) \end{array}

We can then use the points to calculate the slope.

\begin{array}{l} \begin{array}{l} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{array} \\ = \frac{1 - (- 2)}{8 - 3} \\ = \frac{3}{5} \end{array}

Substitute the slope and the coordinates of one of the points into the point-slope form.

\begin{array}{l} y - y_{1} = m (x - x_{1}) \\ y - (- 2) = \frac{3}{5} (x - 3) \end{array}

We can use algebra to rewrite the equation in the slope-intercept form.

\begin{array}{l} y + 2 = \frac{3}{5} (x - 3) \\ y + 2 = \frac{3}{5} x - \frac{9}{5} \\ y = \frac{3}{5} x - \frac{19}{5} \end{array}

Modeling Real-World Problems with Linear Functions

In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.

Example 10

Using a Linear Function to Determine the Number of Songs in a Music Collection

Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, $N,$ in his collection as a function of time, $t,$ the number of months. How many songs will he own in a year?

Solution

The initial value for this function is 200 because he currently owns 200 songs, so $N (0) = 200,$ which means that $b = 200.$

The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that $m = 15.$ We can substitute the initial value and the rate of change into the slope-intercept form of a line.

An image illustrating the mapping of numerical values to the parameters of a general linear function. The equation f(x) = mx + b is shown at the top. An arrow points from the number 15 to 'm', and another arrow points from the number 200 to 'b'. Below this, the specific linear function N(t) = 15t + 200 is displayed, showing how 15 replaces 'm' (slope) and 200 replaces 'b' (y-intercept) from the general form.

We can write the formula $N (t) = 15 t + 200.$

With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at $t = 12.$

\begin{array}{l} N (12) = 15 (12) + 200 \\ = 180 + 200 \\ = 380 \end{array}

Marcus will have 380 songs in 12 months.

Analysis

Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.

Example 11

Using a Linear Function to Calculate Salary Plus Commission

Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, 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.

Solution

The given information gives us two input-output pairs: $(3,760)$ and $(5,920) .$ We start by finding the rate of change.

\begin{array}{l} m = \frac{920 - 760}{5 - 3} \\ = \frac{$160}{2 policies} \\ = $80 per policy \end{array}

Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.

We can then solve for the initial value.

\begin{array}{l} I (n) = 80 n + b \\ 760 = 80 (3) + b & when n = 3, I (3) = 760 \\ 760 - 80 (3) = b \\ 520 = b \end{array}

The value of $b$ is the starting value for the function and represents Ilya’s income when $n = 0,$ or when 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.

We can now write the final equation.

I (n) = 80 n + 520

Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.

Example 12

Using Tabular Form to Write an Equation for a Linear Function

Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.

Table 1 Two rows and five columns. The first row is labeled, 'w, the numers of weeks'. The second row is labeled is labeled, 'P(w), number of rats'. Reading the remaining rows as ordered pairs (i.e., (w, P(w)), we have the following values: (0, 1000), (2, 1080), (4, 1160), and (6, 1240).
w, number of weeks	0	2	4	6
P(w), number of rats	1000	1080	1160	1240

Solution

We can see from the table that the initial value for the number of rats is 1000, so $b = 1000.$

Rather than solving for $m,$ we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.

P (w) = 40 w + 1000

If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using $(2, 1080)$ and $(6, 1240)$

\begin{array}{l} m = \frac{1240 - 1080}{6 - 2} \\ = \frac{160}{4} \\ = 40 \end{array}

Try It #6

A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet, $x$ months since the measurements began. Write a linear function, $H (x),$ where $x$ is the number of months since the start of the experiment.

Table 2 Two rows and six columns. The first row is labeled, 'x'. The second row is labeled is labeled, 'H(x)'. Reading the remaining rows as ordered pairs (i.e., (x, H(x)), we have the following values: (0, 12.5), (2, 13.5), (4, 14.5), (8, 16.5), and (12, 18.5).
$x$	0	2	4	8	12
$H (x)$	12.5	13.5	14.5	16.5	18.5

Solution

$H (x) = 0.5 x + 12.5$

Key Equations

...
slope-intercept form of a line	$f (x) = m x + b$
slope	$m = \frac{change in output (rise)}{change in input (run)} = \frac{Δ y}{Δ x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
point-slope form of a line	$y - y_{1} = m (x - x_{1})$

Key Concepts

The ordered pairs given by a linear function represent points on a line.
Linear functions can be represented in words, function notation, tabular form, and graphical form. See Example 1.
The rate of change of a linear function is also known as the slope.
An equation in the slope-intercept form of a line includes the slope and the initial value of the function.
The initial value, or y-intercept, is the output value when the input of a linear function is zero. It is the y-value of the point at which the line crosses the y-axis.
An increasing linear function results in a graph that slants upward from left to right and has a positive slope.
A decreasing linear function results in a graph that slants downward from left to right and has a negative slope.
A constant linear function results in a graph that is a horizontal line.
Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or constant. See Example 2.
The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line. See Example 3 and Example 4.
The slope and initial value can be determined given a graph or any two points on the line.
One type of function notation is the slope-intercept form of an equation.
The point-slope form is useful for finding a linear equation when given the slope of a line and one point. See Example 5.
The point-slope form is also convenient for finding a linear equation when given two points through which a line passes. See Example 6.
The equation for a linear function can be written if the slope $m$ and initial value $b$ are known. See Example 7, Example 8, and Example 9.
A linear function can be used to solve real-world problems. See Example 10 and Example 11.
A linear function can be written from tabular form. See Example 12.

Section Exercises

Verbal

Exercise 1

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.

Solution

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

Exercise 2

Maria is climbing a mountain. Maria's elevation, $E (t),$ in feet after $t$ minutes is given by $E (t) = 1200 + 40 t .$ Write a complete sentence describing Maria’s starting elevation and how it is changing over time.

Exercise 3

Jessica 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 in miles per hour?

Solution

3 miles per hour

Exercise 4

Sonya is currently 10 miles from home and is walking farther away at 2 miles per hour. Write an equation for her distance from home t hours from now.

Exercise 5

A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.

Solution

$d (t) = 100 - 10 t$

Exercise 6

Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left after riding $n$ rides?

Algebraic

For the following exercises, determine whether the equation of the curve can be written as a linear function.

Exercise 7

$y = \frac{1}{4} x + 6$

Solution

Yes.

Exercise 8

$y = 3 x - 5$

Exercise 9

$y = 3 x^{2} - 2$

Solution

No.

Exercise 10

$3 x + 5 y = 15$

Exercise 11

$3 x^{2} + 5 y = 15$

Solution

No.

Exercise 12

$3 x + 5 y^{2} = 15$

Exercise 13

$- 2 x^{2} + 3 y^{2} = 6$

Solution

No.

Exercise 14

$- \frac{x - 3}{5} = 2 y$

For the following exercises, determine whether each function is increasing or decreasing.

Exercise 15

$f (x) = 4 x + 3$

Solution

Increasing.

Exercise 16

$g (x) = 5 x + 6$

Exercise 17

$a (x) = 5 - 2 x$

Solution

Decreasing.

Exercise 18

$b (x) = 8 - 3 x$

Exercise 19

$h (x) = - 2 x + 4$

Solution

Decreasing.

Exercise 20

$k (x) = - 4 x + 1$

Exercise 21

$j (x) = \frac{1}{2} x - 3$

Solution

Increasing.

Exercise 22

$p (x) = \frac{1}{4} x - 5$

Exercise 23

$n (x) = - \frac{1}{3} x - 2$

Solution

Decreasing.

Exercise 24

$m (x) = - \frac{3}{8} x + 3$

For the following exercises, find the slope of the line that passes through the two given points.

Exercise 25

$(2, 4)$ and $(4, 10)$

Solution

3

Exercise 26

$(1, 5)$ and $(4, 11)$

Exercise 27

$(−1, 4)$ and $(5, 2)$

Solution

$- \frac{1}{3}$

Exercise 28

$(8, −2)$ and $(4, 6)$

Exercise 29

$(6, 11)$ and $(- 4, 3)$

Solution

$\frac{4}{5}$

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

Exercise 30

$f (- 5) = - 4,$ and $f (5) = 2$

Exercise 31

$f (−1) = 4$ and $f (5) = 1$

Solution

$f (x) = - \frac{1}{2} x + \frac{7}{2}$

Exercise 32

$(2, 4)$ and $(4, 10)$

Exercise 33

Passes through $(1, 5)$ and $(4, 11)$

Solution

$y = 2 x + 3$

Exercise 34

Passes through $(- 1, 4)$ and $(5, 2)$

Exercise 35

Passes through $(- 2, 8)$ and $(4, 6)$

Solution

$y = - \frac{1}{3} x + \frac{22}{3}$

Exercise 36

x intercept at $(- 2, 0)$ and y intercept at $(0, −3)$

Exercise 37

x intercept at $(- 5, 0)$ and y intercept at $(0, 4)$

Solution

$y = \frac{4}{5} x + 4$

Graphical

For the following exercises, find the slope of the lines graphed.

Exercise 38

A graph shows a straight line on a Cartesian coordinate plane. The x-axis ranges from -6 to 6, and the y-axis from -6 to 6. The line passes through (-2, 0) and (0, 1) and has a positive slope.

Exercise 39

A coordinate plane displays a straight line. The x-axis is labeled from -6 to 6, and the y-axis is labeled from -6 to 6. The line has a negative slope, descending from left to right. It intersects the y-axis at the point (0, 5) and the x-axis at the point (4, 0). The line passes through points such as (0, 5), (4, 0) and extends infinitely in both directions as indicated by arrows at its ends.

Solution

$- \frac{5}{4}$

Exercise 40

A horizontal line is plotted on a coordinate plane, positioned at y = -2.

For the following exercises, write an equation for the lines graphed.

Exercise 41

A Cartesian coordinate system is shown with both the x-axis and y-axis ranging from -6 to 6. A straight blue line is plotted, extending across the graph from the bottom-left to the top-right. The line passes through several grid points, including (-2, 0) on the x-axis, (0, 1) on the y-axis, and (2, 2). The line has a positive slope, rising one unit for every two units moved to the right.

Solution

$y = \frac{2}{3} x + 1$

Exercise 42

A Cartesian coordinate system shows a straight blue line. The x-axis ranges from -6 to 6, and the y-axis ranges from -6 to 6. The line passes through the y-axis at (0, 5) and the x-axis at (4, 0). It slopes downwards from left to right, indicating a negative slope, and extends indefinitely in both directions, as shown by arrows.

Exercise 43

A Cartesian coordinate system graph displays a downward-sloping blue line. The line intersects the y-axis at (0, 3) and the x-axis at roughly (1.5, 0), extending infinitely.

Solution

$y = - 2 x + 3$

Exercise 44

A Cartesian coordinate plane with x and y axes ranging from -6 to 6. A blue straight line is plotted, representing the equation y = 4x - 2. The line passes through the y-intercept at (0, -2) and has a positive slope, passing through points such as (1, 2) and (2, 6).

Exercise 45

A Cartesian coordinate system is shown with an x-axis labeled from -6 to 6 and a y-axis labeled from -6 to 6. A solid blue horizontal line is drawn at y=3, extending across the entire visible x-axis with arrows on both ends, indicating it continues infinitely in both directions.

Solution

$y = 3$

Exercise 46

A graph displays a horizontal line at y = -2.5 on a coordinate plane, extending from x = -6 to x = 6 across the grid.

Numeric

For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.

Exercise 47

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'g(x)'. Reading the remaining rows as ordered pairs (i.e., (x, g(x)), we have the following values: (0, 5), (5, -10), (10, -25), and (15, -40).
$x$	0	5	10	15
$g (x)$	5	–10	–25	–40

Solution

Linear, $g (x) = - 3 x + 5$

Exercise 48

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'h(x)'. Reading the remaining rows as ordered pairs (i.e., (x, h(x)), we have the following values: (0, 5), (5, 30), (10, 105), and (15, 230).
$x$	0	5	10	15
$h (x)$	5	30	105	230

Exercise 49

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'f(x)'. Reading the remaining rows as ordered pairs (i.e., (x, f(x)), we have the following values: (0,- 5), (5, 20), (10, 45), and (15, 70).
$x$	0	5	10	15
$f (x)$	–5	20	45	70

Solution

Linear, $f (x) = 5 x - 5$

Exercise 50

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'k(x)'. Reading the remaining rows as ordered pairs (i.e., (x, k(x)), we have the following values: (5, 13), (10, 28), (20, 58), and (25, 73).
$x$	5	10	20	25
$k (x)$	13	28	58	73

Exercise 51

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'g(x)'. Reading the remaining rows as ordered pairs (i.e., (x, g(x)), we have the following values: (0, 6), (5, -10), (10, -25), and (15, -40).
$x$	0	2	4	6
$g (x)$	6	–19	–44	–69

Solution

Linear, $g (x) = - \frac{25}{2} x + 6$

Exercise 52

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'h(x)'. Reading the remaining rows as ordered pairs (i.e., (x, h(x)), we have the following values: (2, 13), (4, 23), (8, 43), and (10, 53).
$x$	2	4	6	8
$f (x)$	–4	16	36	56

Exercise 53

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'f(x)'. Reading the remaining rows as ordered pairs (i.e., (x, f(x)), we have the following values: (2, -4), (4, 16), (6, 36), and (8, 56).
$x$	2	4	6	8
$f (x)$	–4	16	36	56

Solution

Linear, $f (x) = 10 x - 24$

Exercise 54

Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'k(x)'. Reading the remaining rows as ordered pairs (i.e., (x, k(x)), we have the following values: (0, 6), (2, 31), (6, 106), and (8, 231).
$x$	0	2	6	8
$k (x)$	6	31	106	231

Technology

Exercise 55

If $f$ is a linear function, $f (0.1) = 11.5, and f (0.4) = - 5.9,$ find an equation for the function.

Solution

$f (x) = - 58 x + 17.3$

Exercise 56

Graph the function $f$ on a domain of $[- 10, 10] : f (x) = 0.02 x - 0.01.$ Enter the function in a graphing utility. For the viewing window, set the minimum value of $x$ to be $- 10$ and the maximum value of $x$ to be $10.$

Exercise 57

Graph the function $f$ on a domain of $[- 10, 10] : f x) = 2, 500 x + 4, 000$

Solution

A graph shows a straight line labeled 'f' on a coordinate plane. The x-axis ranges from -8 to 8 in increments of 1, and the y-axis ranges from -20000 to 25000 in increments of 5000. The line 'f' passes through the origin (0,0) and has a positive slope. Notable points on the line include (-4, -10000) and (4, 10000).

Exercise 58

Table 3 shows the input, $w,$ and output, $k,$ for a linear function $k .$ a. Fill in the missing values of the table. b. Write the linear function $k,$ round to 3 decimal places.

Table 3 Two rows and five columns. The first row is labeled, 'w'. The second row is labeled is labeled, 'k'. Reading the remaining rows as ordered pairs (i.e., (w, k), we have the following values: (-10, 30), (5.5, -26), (67.5, a), and (b, -44).
$w$	–10	5.5	67.5	b
$k$	30	–26	a	–44

Exercise 59

Table 4 shows the input, $p,$ and output, $q,$ for a linear function $q .$ a. Fill in the missing values of the table. b. Write the linear function $k .$

Table 4 Two rows and five columns. The first row is labeled, 'w'. The second row is labeled is labeled, 'k'. Reading the remaining rows as ordered pairs (i.e., (p, q), we have the following values: (0.5, 400), (0.8, 700), (12, a), and (b, 1,000,000).
$p$	0.5	0.8	12	b
$q$	400	700	a	1,000,000

Solution

a. $a = 11, 900$ ; $b = 1000.1$ b. $q (p) = 1000 p - 100$

Exercise 60

Graph the linear function $f$ on a domain of $[- 10, 10]$ for the function whose slope is $\frac{1}{8}$ and y-intercept is $\frac{31}{16}$ . Label the points for the input values of $- 10$ and $10.$

Exercise 61

Graph the linear function $f$ on a domain of $[- 0.1, 0.1]$ for the function whose slope is 75 and y-intercept is $- 22.5$ . Label the points for the input values of $- 0.1$ and $0.1.$

Solution

A 2D graph displays an x-axis from -0.1 to 0.1 and a y-axis from 10 to -30. A blue line segment ascends from approximately (-0.09, -30) to (0.09, -15).

Exercise 62

Graph the linear function

f

where

f (x) = a x + b

on the same set of axes on a domain of

[- 4, 4]

for the following values of

a

and

b .

$a = 2; b = 3$
$a = 2; b = 4$
$a = 2; b = - 4$
$a = 2; b = - 5$

Extensions

Exercise 63

Find the value of $x$ if a linear function goes through the following points and has the following slope: $(x, 2), (- 4, 6), m = 3$

Solution

$x = - \frac{16}{3}$

Exercise 64

Find the value of y if a linear function goes through the following points and has the following slope: $(10, y), (25, 100), m = - 5$

Exercise 65

Find the equation of the line that passes through the following points: $(a, b)$ and $(a, b + 1)$

Solution

$x = a$

Exercise 66

Find the equation of the line that passes through the following points: $(2 a, b)$ and $(a, b + 1)$

Exercise 67

Find the equation of the line that passes through the following points: $(a, 0)$ and $(c, d)$

Solution

$y = \frac{d}{c - a} x - \frac{a d}{c - a}$

Real-World Applications

Exercise 68

At noon, a barista notices that they have $20 in their tip jar. If the barista makes an average of $0.50 from each customer, how much will they have in the tip jar if they serve $n$ more customers during the shift?

Exercise 69

A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session?

Solution

$45 per training session.

Exercise 70

A clothing business finds there is a linear relationship between the number of shirts, $n,$ it can sell and the price, $p,$ it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $$ 30$ , while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form $p (n) = m n + b$ that gives the price $p$ they can charge for $n$ shirts.

Exercise 71

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.

Solution

The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.

Exercise 72

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 relationship in the form $y = m n + b$ that gives the yield when $n$ stalks are planted.

Exercise 73

A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.

Solution

The slope is $- 400.$ This means for every year between 1960 and 1989, the population dropped by 400 per year in the city.

Exercise 74

A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, $P (t),$ for the population $t$ years after 2003.

Exercise 75

Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: $I (x) = 1054 x + 23, 286,$ where $x$ is the number of years after 1990. Which of the following interprets the slope in the context of the problem?

As of 1990, average annual income was $23,286.
In the ten-year period from 1990–1999, average annual income increased by a total of $1,054.
Each year in the decade of the 1990s, average annual income increased by $1,054.
Average annual income rose to a level of $23,286 by the end of 1999.

Solution

c.

Exercise 76

When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of $C,$ the Celsius temperature, $F (C) .$

Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius.
Find and interpret $F (28) .$
Find and interpret $F (–40) .$