College Algebra with Corequisite Support

# 4.1Linear Functions

### Corequisite Skills

#### Learning Objectives

• Find the slope of a line (IA 3.2.1)
• Find an equation of the line given two points (IA 3.3.3)

#### Objective 1: Find the slope of a line. (IA 3.2.1)

Linear functions are a specific type of function that can be used to model many real-world applications, such as the growth of a plant, earned salary, the distance a train travels over time, or the costs to start a new business. In this section, we will explore linear functions, their graphs, and how to find them using data points.

##### Linear Function

A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line

$f(x)=mx+bf(x)=mx+b$

where $bb$ is the initial or starting value of the function (when input, $x=0x=0$), and $mm$ is the constant rate of change, or slope of the function. The $yy$ -intercept is at ( $0,b0,b$ ).

When interpreting slope, it will be important to consider the units of measurement. Make sure to always attach these units to both the numerator and denominator when they are provided to you.

### Quick Guide to Slopes of Lines ### Example 1

Find the slope of the line shown. ##### Practice Makes Perfect

Find the slope of the line.

1.

Find the slope of the line below: 2.

Find the slope of the following line: $y=3y=3$

3.

Find the slope of the following line: $x=-2x=-2$

4.

Find the slope of the following line: $y=-5x-6y=-5x-6$

5.

Find the slope of the following line: $2x-4y=52x-4y=5$

6.

Use the slope formula to find the slope of the line between the following pair of points. $(-2,4),(3,-1)(-2,4),(3,-1)$

7.

Use the slope formula to find the slope of the linear function satisfying the condition below. $f(-2)=-1,f(6)=5f(-2)=-1,f(6)=5$

#### Objective 2: Find an equation of the line given two points. (IA 3.3.3)

##### Find an Equation of the Line Given Two Points

When data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when two points are given by following the steps below.

### How To

1. Step 1. Find the slope using the given points.
2. Step 2. Choose one point and label its coordinates $(x1,y1)(x1,y1)$ .
3. Step 3. Plug and $y1y1$ into point-slope form, $y-y1=m(x-x1)y-y1=m(x-x1)$ .
4. Step 4. Write the equation in slope-intercept form, $y=mx+by=mx+b$ .

### Example 2

Find an equation of the line given two points.

Find the equation of a line containing the points (−4, −3) and (1, −5)

##### Practice Makes Perfect

Find an equation of the line given two points.

8.

Find the equation of a line containing the given points. Write the equation in slope-intercept form. $(4,3)(4,3)$ and $(8,1)(8,1)$

9.

Find the equation of a line containing the given points. Write the equation in slope-intercept form. $(-5,-3)(-5,-3)$ and $(4,-6)(4,-6)$

10.

Find the equation of a line containing the given points. Write the equation in slope-intercept form. $f(-2)=8,f(4)=6f(-2)=8,f(4)=6$

11.

Derek notices the amount he receives in tips each night is a linear function of the number of tables he waits on. On Friday evening he waits on 22 tables and receives $87 in tips, and on Tuesday evening he waits on 18 tables and receives$73 in tips.

1. Record the information given above as two data points.
2. Find the slope of this linear function in terms of $tabletable$ .

### Interpreting Slope as a Rate of Change

In the examples we have seen so far, the slope was provided to us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input, $x 1 x 1$ and $x 2 , x 2 ,$ and two corresponding values for the output, $y 1 y 1$ and $y 2 y 2$ —which can be represented by a set of points, and —we can calculate the slope $m. m.$

Note that in function notation we can obtain two corresponding values for the output $y 1 y 1$ and $y 2 y 2$ for the function $f, f,$ $y 1 =f( x 1 ) y 1 =f( x 1 )$ and $y 2 =f( x 2 ), y 2 =f( x 2 ),$ so we could equivalently write

$m= f( x 2 )–f( x 1 ) x 2 – x 1 m= f( x 2 )–f( x 1 ) x 2 – x 1$

Figure 6 indicates how the slope of the line between the points, $( x 1 , y 1 ) ( x 1 , y 1 )$ and $( x 2 , y 2 ), ( x 2 , y 2 ),$ is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the slant is.

Figure 6 The slope of a function is calculated by the change in $y y$ divided by the change in $x. x.$ It does not matter which coordinate is used as the $( x 2, y 2 ) ( x 2, y 2 )$ and which is the $( x 1 , y 1 ), ( x 1 , y 1 ),$ as long as each calculation is started with the elements from the same coordinate pair.

### Q&A

Are the units for slope always

Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.

### Calculate Slope

The slope, or rate of change, of a function $m m$ can be calculated according to the following:

where $x 1 x 1$ and $x 2 x 2$ are input values, $y 1 y 1$ and $y 2 y 2$ are output values.

### How To

Given two points from a linear function, calculate and interpret the slope.

1. Determine the units for output and input values.
2. Calculate the change of output values and change of input values.
3. Interpret the slope as the change in output values per unit of the input value.

### Example 3

#### Finding the Slope of a Linear Function

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

#### 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. Note that if we had reversed them, we would have obtained the same slope.

$m= ( −2 )−(1) 3−8 = −3 −5 = 3 5 m= ( −2 )−(1) 3−8 = −3 −5 = 3 5$
Try It #1

If $f(x) f(x)$ is a linear function, and $(2,3) (2,3)$ and $(0,4) (0,4)$ are points on the line, find the slope. Is this function increasing or decreasing?

### 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.

#### 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.

Try It #2

The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012.

### Writing and Interpreting an Equation for a Linear Function

Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form. Now we can choose which method to use to write equations for linear functions 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 $f f$ in Figure 7.

Figure 7

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 $(0,7) (0,7)$ and $(4,4). (4,4).$

$m = y 2 − y 1 x 2 − x 1 = 4−7 4−0 = − 3 4 m = y 2 − y 1 x 2 − x 1 = 4−7 4−0 = − 3 4$

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

$y− y 1 = m(x− x 1 ) y−4 = − 3 4 (x−4) y− y 1 = m(x− x 1 ) y−4 = − 3 4 (x−4)$

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

$y−4 = − 3 4 (x−4) y−4 = − 3 4 x+3 y = − 3 4 x+7 y−4 = − 3 4 (x−4) y−4 = − 3 4 x+3 y = − 3 4 x+7$

If we want 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, $b=7. b=7.$ We now have the initial value $b b$ and the slope $m m$ so we can substitute $m m$ and $b b$ into the slope-intercept form of a line. So the function is $f(x)=− 3 4 x+7, f(x)=− 3 4 x+7,$ and the linear equation would be $y=− 3 4 x+7. y=− 3 4 x+7.$

### How To

Given the graph of a linear function, write an equation to represent the function.

1. Identify two points on the line.
2. Use the two points to calculate the slope.
3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection.
4. Substitute the slope and y-intercept into the slope-intercept form of a line equation.

### Example 5

#### Writing an Equation for a Linear Function

Write an equation for a linear function given a graph of $f f$ shown in Figure 8.

Figure 8

#### Analysis

This makes sense because we can see from Figure 9 that the line crosses the y-axis at the point $( 0,2 ), ( 0,2 ),$ which is the y-intercept, so $b=2. b=2.$

Figure 9

### Example 6

#### 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 $C C$ where $C( x ) C( x )$ is the cost for $x x$ items produced in a given month.

#### Analysis

If Ben produces 100 items in a month, his monthly cost is found by substituting 100 for $x. x.$

$C(100) = 1250+37.5(100) = 5000 C(100) = 1250+37.5(100) = 5000$

So his monthly cost would be $5,000. ### Example 7 #### Writing an Equation for a Linear Function Given Two Points If $f f$ is a linear function, with $f(3)=−2, f(3)=−2,$ and $f(8)=1, f(8)=1,$ find an equation for the function in slope-intercept form. Try It #3 If $f(x) f(x)$ is a linear function, with $f(2)=–11, f(2)=–11,$ and $f(4)=−25, f(4)=−25,$ write an equation for the function in slope-intercept form. ### 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. ### How To Given a linear function $f f$ and the initial value and rate of change, evaluate $f( c ). f( c ).$ 1. Determine the initial value and the rate of change (slope). 2. Substitute the values into $f(x)=mx+b. f(x)=mx+b.$ 3. Evaluate the function at $x=c. x=c.$ ### Example 8 #### 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, $N, N,$ in his collection as a function of time, $t, t,$ the number of months. How many songs will he own at the end of one year? #### 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 9 #### Using a Linear Function to Calculate Salary Based on Commission Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income $I, I,$ depends on the number of new policies, $n, n,$ 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 $I(n), I(n),$ and interpret the meaning of the components of the equation. ### Example 10 #### 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.  number of weeks, w 0 2 4 6 number of rats, P(w) 1000 1080 1160 1240 Table 1 ### Q&A Is the initial value always provided in a table of values like Table 1? No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into $f(x)=mx+b, f(x)=mx+b,$ and solve for $b. b.$ Try It #4 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, $x x$ months since the measurements began. Write a linear function, $H(x), H(x),$ where $x x$ is the number of months since the start of the experiment.  x 0 2 4 8 12 H(x) 12.5 13.5 14.5 16.5 18.5 Table 2 ### Graphing Linear Functions Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third method is by using transformations of the identity function $f(x)=x. f(x)=x.$ #### Graphing a Function by Plotting Points To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, $f(x)=2x, f(x)=2x,$ we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point $(1,2). (1,2).$ Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point $(2,4). (2,4).$ Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. ### How To Given a linear function, graph by plotting points. 1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points. ### Example 11 #### Graphing by Plotting Points Graph $f(x)=− 2 3 x+5 f(x)=− 2 3 x+5$ by plotting points. #### Analysis The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function. ### Try It #5 Graph $f(x)=− 3 4 x+6 f(x)=− 3 4 x+6$ by plotting points. #### Graphing a Function Using y-intercept and Slope Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set $x=0 x=0$ in the equation. The other characteristic of the linear function is its slope. Let’s consider the following function. $f(x)= 1 2 x+1 f(x)= 1 2 x+1$ The slope is $1 2 . 1 2 .$ Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is the point on the graph when $x=0. x=0.$ The graph crosses the y-axis at $(0,1). (0,1).$ Now we know the slope and the y-intercept. We can begin graphing by plotting the point $(0,1). (0,1).$ We know that the slope is the change in the y-coordinate over the change in the x-coordinate. This is commonly referred to as rise over run, $m= rise run . m= rise run .$ From our example, we have $m= 1 2 , m= 1 2 ,$ which means that the rise is 1 and the run is 2. So starting from our y-intercept $(0,1), (0,1),$ we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 12. Figure 12 ### Graphical Interpretation of a Linear Function In the equation $f(x)=mx+b f(x)=mx+b$ • $b b$ is the y-intercept of the graph and indicates the point $(0,b) (0,b)$ at which the graph crosses the y-axis. • $m m$ is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope: ### Q&A Do all linear functions have y-intercepts? Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line is parallel to the y-axis does not have a y-intercept, but it is not a function.) ### How To Given the equation for a linear function, graph the function using the y-intercept and slope. 1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept. 4. Use $rise run rise run$ to determine at least two more points on the line. 5. Sketch the line that passes through the points. ### Example 12 #### Graphing by Using the y-intercept and Slope Graph $f(x)=− 2 3 x+5 f(x)=− 2 3 x+5$ using the y-intercept and slope. #### Analysis The graph slants downward from left to right, which means it has a negative slope as expected. ### Try It #6 Find a point on the graph we drew in Example 12 that has a negative x-value. #### Graphing a Function Using Transformations Another option for graphing is to use a transformation of the identity function $f(x)=x. f(x)=x.$ A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression. ##### Vertical Stretch or Compression In the equation $f(x)=mx, f(x)=mx,$ the $m m$ is acting as the vertical stretch or compression of the identity function. When $m m$ is negative, there is also a vertical reflection of the graph. Notice in Figure 14 that multiplying the equation of $f(x)=x f(x)=x$ by $m m$ stretches the graph of $f f$ by a factor of $m m$ units if $m>1 m>1$ and compresses the graph of $f f$ by a factor of $m m$ units if $0 This means the larger the absolute value of $m, m,$ the steeper the slope. Figure 14 Vertical stretches and compressions and reflections on the function $f(x)=x f(x)=x$ ##### Vertical Shift In $f(x)=mx+b, f(x)=mx+b,$ the $b b$ acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in Figure 15 that adding a value of $b b$ to the equation of $f( x )=x f( x )=x$ shifts the graph of $f f$ a total of $b b$ units up if $b b$ is positive and $|b| |b|$ units down if $b b$ is negative. Figure 15 This graph illustrates vertical shifts of the function $f(x)=x. f(x)=x.$ 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 to graph this type of function, it is still important to practice each method. ### How To Given the equation of a linear function, use transformations to graph the linear function in the form $f( x )=mx+b. f( x )=mx+b.$ 1. Graph $f( x )=x. f( x )=x.$ 2. Vertically stretch or compress the graph by a factor $m. m.$ 3. Shift the graph up or down $b b$ units. ### Example 13 #### Graphing by Using Transformations Graph $f(x)= 1 2 x−3 f(x)= 1 2 x−3$ using transformations. ### Try It #7 Graph $f(x)=4+2x f(x)=4+2x$ using transformations. ### Q&A In Example 15, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2. $f(2) = 1 2 (2)−3 = 1−3 = −2 f(2) = 1 2 (2)−3 = 1−3 = −2$ ### Writing the Equation for a Function from the Graph of a Line Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 18. We can see right away that the graph crosses the y-axis at the point $(0,4) (0,4)$ so this is the y-intercept. Figure 18 Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point $(–2,0). (–2,0).$ To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be $m= rise run = 4 2 =2 m= rise run = 4 2 =2$ Substituting the slope and y-intercept into the slope-intercept form of a line gives $y=2x+4 y=2x+4$ ### How To Given a graph of linear function, find the equation to describe the function. 1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute the y-intercept and slope into the slope-intercept form of a line. ### Example 14 #### Matching Linear Functions to Their Graphs Match each equation of the linear functions with one of the lines in Figure 19. Figure 19 #### Finding the x-intercept of a Line So far we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis. Recall that a function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero. To find the x-intercept, set a function $f(x) f(x)$ equal to zero and solve for the value of $x. x.$ For example, consider the function shown. $f(x)=3x−6 f(x)=3x−6$ Set the function equal to 0 and solve for $x. x.$ $0 = 3x−6 6 = 3x 2 = x x = 2 0 = 3x−6 6 = 3x 2 = x x = 2$ The graph of the function crosses the x-axis at the point $(2,0). (2,0).$ ### Q&A Do all linear functions have x-intercepts? No. However, linear functions of the form $y=c, y=c,$ where $c c$ is a nonzero real number are the only examples of linear functions with no x-intercept. For example, $y=5 y=5$ is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 21. Figure 21 ### x-intercept The x-intercept of the function is value of $x x$ when $f(x)=0. f(x)=0.$ It can be solved by the equation $0=mx+b. 0=mx+b.$ ### Example 15 #### Finding an x-intercept Find the x-intercept of $f(x)= 1 2 x−3. f(x)= 1 2 x−3.$ #### Analysis A graph of the function is shown in Figure 22. We can see that the x-intercept is $(6,0) (6,0)$ as we expected. Figure 22 ### Try It #8 Find the x-intercept of $f(x)= 1 4 x−4. f(x)= 1 4 x−4.$ #### Describing Horizontal and Vertical Lines There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 23, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use $m=0 m=0$ in the equation $f(x)=mx+b, f(x)=mx+b,$ the equation simplifies to $f(x)=b. f(x)=b.$ In other words, the value of the function is a constant. This graph represents the function $f(x)=2. f(x)=2.$ Figure 23 A horizontal line representing the function $f(x)=2 f(x)=2$ A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined. Figure 24 Example of how a line has a vertical slope. 0 in the denominator of the slope. A vertical line, such as the one in Figure 25, has an x-intercept, but no y-intercept unless it’s the line $x=0. x=0.$ This graph represents the line $x=2. x=2.$ Figure 25 The vertical line, $x=2, x=2,$ which does not represent a function ### Horizontal and Vertical Lines Lines can be horizontal or vertical. A horizontal line is a line defined by an equation in the form $f(x)=b. f(x)=b.$ A vertical line is a line defined by an equation in the form $x=a. x=a.$ ### Example 16 #### Writing the Equation of a Horizontal Line Write the equation of the line graphed in Figure 26. Figure 26 ### Example 17 #### Writing the Equation of a Vertical Line Write the equation of the line graphed in Figure 27. Figure 27 ### Determining Whether Lines are Parallel or Perpendicular The two lines in Figure 28 are parallel lines: they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the other, they would become coincident. Figure 28 Parallel lines We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 29 are perpendicular. Figure 29 Perpendicular lines Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is $1. 1.$ So, if are negative reciprocals of one another, they can be multiplied together to yield $–1. –1.$ $m 1 m 2 =−1 m 1 m 2 =−1$ To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is $1 8 , 1 8 ,$ and the reciprocal of $1 8 1 8$ is 8. To find the negative reciprocal, first find the reciprocal and then change the sign. As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular. The product of the slopes is –1. $−4( 1 4 )=−1 −4( 1 4 )=−1$ ### Parallel and Perpendicular Lines Two lines are parallel lines if they do not intersect. The slopes of the lines are the same. If and only if $b 1 = b 2 b 1 = b 2$ and $m 1 = m 2 , m 1 = m 2 ,$ we say the lines coincide. Coincident lines are the same line. Two lines are perpendicular lines if they intersect to form a right angle. $m 1 m 2 =−1,so m 2 =− 1 m 1 m 1 m 2 =−1,so m 2 =− 1 m 1$ ### Example 18 #### Identifying Parallel and Perpendicular Lines Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines. $f(x) = 2x+3 h(x) = −2x+2 g(x) = 1 2 x−4 j(x) = 2x−6 f(x) = 2x+3 h(x) = −2x+2 g(x) = 1 2 x−4 j(x) = 2x−6$ #### Analysis A graph of the lines is shown in Figure 30. Figure 30 The graph shows that the lines $f(x)=2x+3 f(x)=2x+3$ and $j(x)=2x–6 j(x)=2x–6$ are parallel, and the lines $g(x)= 1 2 x–4 g(x)= 1 2 x–4$ and $h(x)=−2x+2 h(x)=−2x+2$ are perpendicular. ### Writing the Equation of a Line Parallel or Perpendicular to a Given Line If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line. #### Writing Equations of Parallel Lines Suppose for example, we are given the equation shown. $f(x)=3x+1 f(x)=3x+1$ We know that the slope of the line formed by the function is 3. We also know that the y-intercept is $(0,1). (0,1).$ Any other line with a slope of 3 will be parallel to $f(x). f(x).$ So the lines formed by all of the following functions will be parallel to $f(x). f(x).$ $g(x) = 3x+6 h(x) = 3x+1 p(x) = 3x+ 2 3 g(x) = 3x+6 h(x) = 3x+1 p(x) = 3x+ 2 3$ Suppose then we want to write the equation of a line that is parallel to $f f$ and passes through the point $(1,7). (1,7).$ This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value of $b b$ will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form. $y− y 1 = m(x− x 1 ) y−7 = 3(x−1) y−7 = 3x−3 y = 3x+4 y− y 1 = m(x− x 1 ) y−7 = 3(x−1) y−7 = 3x−3 y = 3x+4$ So $g(x)=3x+4 g(x)=3x+4$ is parallel to $f( x )=3x+1 f( x )=3x+1$ and passes through the point $(1,7). (1,7).$ ### How To Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point. 1. Find the slope of the function. 2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line. 3. Simplify. ### Example 19 #### Finding a Line Parallel to a Given Line Find a line parallel to the graph of $f(x)=3x+6 f(x)=3x+6$ that passes through the point $(3,0). (3,0).$ #### Analysis We can confirm that the two lines are parallel by graphing them. Figure 31 shows that the two lines will never intersect. Figure 31 #### Writing Equations of Perpendicular Lines We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the function shown. $f( x )=2x+4 f( x )=2x+4$ The slope of the line is 2, and its negative reciprocal is $− 1 2 . − 1 2 .$ Any function with a slope of $− 1 2 − 1 2$ will be perpendicular to $f(x). f(x).$ So the lines formed by all of the following functions will be perpendicular to $f(x). f(x).$ $g(x) = − 1 2 x+4 h(x) = − 1 2 x+2 p(x) = − 1 2 x− 1 2 g(x) = − 1 2 x+4 h(x) = − 1 2 x+2 p(x) = − 1 2 x− 1 2$ As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to $f(x) f(x)$ and passes through the point $(4,0). (4,0).$ We already know that the slope is $− 1 2 . − 1 2 .$ Now we can use the point to find the y-intercept by substituting the given values into the slope-intercept form of a line and solving for $b. b.$ $g(x) = mx+b 0 = − 1 2 (4)+b 0 = −2+b 2 = b b = 2 g(x) = mx+b 0 = − 1 2 (4)+b 0 = −2+b 2 = b b = 2$ The equation for the function with a slope of $− 1 2 − 1 2$ and a y-intercept of 2 is $g(x)=− 1 2 x+2 g(x)=− 1 2 x+2$ So $g(x)=− 1 2 x+2 g(x)=− 1 2 x+2$ is perpendicular to $f( x )=2x+4 f( x )=2x+4$ and passes through the point $(4,0). (4,0).$ Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature. ### Q&A A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines? No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted. ### How To Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line. 1. Find the slope of the function. 2. Determine the negative reciprocal of the slope. 3. Substitute the new slope and the values for $x x$ and $y y$ from the coordinate pair provided into $g( x )=mx+b. g( x )=mx+b.$ 4. Solve for $b. b.$ 5. Write the equation of the line. ### Example 20 #### Finding the Equation of a Perpendicular Line Find the equation of a line perpendicular to $f(x)=3x+3 f(x)=3x+3$ that passes through the point $(3,0). (3,0).$ #### Analysis A graph of the two lines is shown in Figure 32. Figure 32 Note that that if we graph perpendicular lines on a graphing calculator using standard zoom, the lines may not appear to be perpendicular. Adjusting the window will make it possible to zoom in further to see the intersection more closely. ### Try It #9 Given the function $h(x)=2x−4, h(x)=2x−4,$ write an equation for the line passing through $( 0,0 ) ( 0,0 )$ that is 1. parallel to $h(x) h(x)$ 2. perpendicular to $h(x) h(x)$ ### How To Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point. 1. Determine the slope of the line passing through the points. 2. Find the negative reciprocal of the slope. 3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values. 4. Simplify. ### Example 21 #### Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point A line passes through the points $(−2,6) (−2,6)$ and $(4,5). (4,5).$ Find the equation of a perpendicular line that passes through the point $(4,5). (4,5).$ ### Try It #10 A line passes through the points, $(−2,−15) (−2,−15)$ and $(2,−3). (2,−3).$ Find the equation of a perpendicular line that passes through the point, $(6,4). (6,4).$ ### Media Access this online resource for additional instruction and practice with linear functions. ### 4.1 Section Exercises #### Verbal 1. Terry is skiing down a steep hill. Terry's elevation, $E(t), E(t),$ in feet after $t t$ seconds is given by $E(t)=3000−70t. E(t)=3000−70t.$ Write a complete sentence describing Terry’s starting elevation and how it is changing over time. 2. 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? 3. 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. 4. If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y-intercepts. 5. If a horizontal line has the equation $f( x )=a f( x )=a$ and a vertical line has the equation $x=a, x=a,$ what is the point of intersection? Explain why what you found is the point of intersection. #### Algebraic For the following exercises, determine whether the equation of the curve can be written as a linear function. 6. $y= 1 4 x+6 y= 1 4 x+6$ 7. $y=3x−5 y=3x−5$ 8. $y=3 x 2 −2 y=3 x 2 −2$ 9. $3x+5y=15 3x+5y=15$ 10. $3 x 2 +5y=15 3 x 2 +5y=15$ 11. $3x+5 y 2 =15 3x+5 y 2 =15$ 12. $−2 x 2 +3 y 2 =6 −2 x 2 +3 y 2 =6$ 13. $− x−3 5 =2y − x−3 5 =2y$ For the following exercises, determine whether each function is increasing or decreasing. 14. $f( x )=4x+3 f( x )=4x+3$ 15. $g( x )=5x+6 g( x )=5x+6$ 16. $a( x )=5−2x a( x )=5−2x$ 17. $b( x )=8−3x b( x )=8−3x$ 18. $h( x )=−2x+4 h( x )=−2x+4$ 19. $k( x )=−4x+1 k( x )=−4x+1$ 20. $j( x )= 1 2 x−3 j( x )= 1 2 x−3$ 21. $p( x )= 1 4 x−5 p( x )= 1 4 x−5$ 22. $n( x )=− 1 3 x−2 n( x )=− 1 3 x−2$ 23. $m( x )=− 3 8 x+3 m( x )=− 3 8 x+3$ For the following exercises, find the slope of the line that passes through the two given points. 24. $(2,4) (2,4)$ and $(4,10) (4,10)$ 25. $(1,5) (1,5)$ and $(4,11) (4,11)$ 26. $(–1,4) (–1,4)$ and $(5,2) (5,2)$ 27. $(8,–2) (8,–2)$ and $(4,6) (4,6)$ 28. $(6,11) (6,11)$ and $(–4,3) (–4,3)$ For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. 29. $f(−5)=−4, f(−5)=−4,$ and $f(5)=2 f(5)=2$ 30. $f(−1)=4, f(−1)=4,$ and $f(5)=1 f(5)=1$ 31. Passes through $(2,4) (2,4)$ and $(4,10) (4,10)$ 32. Passes through $(1,5) (1,5)$ and $(4,11) (4,11)$ 33. Passes through $(−1,4) (−1,4)$ and $(5,2) (5,2)$ 34. Passes through $(−2,8) (−2,8)$ and $(4,6) (4,6)$ 35. x intercept at $(−2,0) (−2,0)$ and y intercept at $(0,−3) (0,−3)$ 36. x intercept at $(−5,0) (−5,0)$ and y intercept at $(0,4) (0,4)$ For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither. 37. $4x−7y=10 7x+4y=1 4x−7y=10 7x+4y=1$ 38. $3y+x=12 −y=8x+1 3y+x=12 −y=8x+1$ 39. $3y+4x=12 −6y=8x+1 3y+4x=12 −6y=8x+1$ 40. $6x−9y=10 3x+2y=1 6x−9y=10 3x+2y=1$ For the following exercises, find the x- and y-intercepts of each equation. 41. $f( x )=−x+2 f( x )=−x+2$ 42. $g( x )=2x+4 g( x )=2x+4$ 43. $h( x )=3x−5 h( x )=3x−5$ 44. $k( x )=−5x+1 k( x )=−5x+1$ 45. $−2x+5y=20 −2x+5y=20$ 46. $7x+2y=56 7x+2y=56$ For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 47. Line 1: Passes through $(0,6) (0,6)$ and $(3,−24) (3,−24)$ Line 2: Passes through $(−1,19) (−1,19)$ and $(8,−71) (8,−71)$ 48. Line 1: Passes through $(−8,−55) (−8,−55)$ and $(10,89) (10,89)$ Line 2: Passes through $(9,−44) (9,−44)$ and $(4,−14) (4,−14)$ 49. Line 1: Passes through $(2,3) (2,3)$ and $(4,−1) (4,−1)$ Line 2: Passes through $(6,3) (6,3)$ and $(8,5) (8,5)$ 50. Line 1: Passes through $(1,7) (1,7)$ and $(5,5) (5,5)$ Line 2: Passes through $(−1,−3) (−1,−3)$ and $(1,1) (1,1)$ 51. Line 1: Passes through $(2,5) (2,5)$ and $(5,−1) (5,−1)$ Line 2: Passes through $(−3,7) (−3,7)$ and $(3,−5) (3,−5)$ For the following exercises, write an equation for the line described. 52. Write an equation for a line parallel to $f( x )=−5x−3 f( x )=−5x−3$ and passing through the point $(2,–12). (2,–12).$ 53. Write an equation for a line parallel to $g(x)=3x−1 g(x)=3x−1$ and passing through the point $(4,9). (4,9).$ 54. Write an equation for a line perpendicular to $h(t)=−2t+4 h(t)=−2t+4$ and passing through the point $(−4,–1). (−4,–1).$ 55. Write an equation for a line perpendicular to $p(t)=3t+4 p(t)=3t+4$ and passing through the point $(3,1). (3,1).$ #### Graphical For the following exercises, find the slope of the line graphed. 56. 57. For the following exercises, write an equation for the line graphed. 58. 59. 60. 61. 62. 63. For the following exercises, match the given linear equation with its graph in Figure 33. Figure 33 64. $f( x )=−x−1 f( x )=−x−1$ 65. $f( x )=−3x−1 f( x )=−3x−1$ 66. $f( x )=− 1 2 x−1 f( x )=− 1 2 x−1$ 67. $f( x )=2 f( x )=2$ 68. $f( x )=2+x f( x )=2+x$ 69. $f( x )=3x+2 f( x )=3x+2$ For the following exercises, sketch a line with the given features. 70. An x-intercept of $(–4,0) (–4,0)$ and y-intercept of $(0,–2) (0,–2)$ 71. An x-intercept $(–2,0) (–2,0)$ and y-intercept of $(0,4) (0,4)$ 72. A y-intercept of $(0,7) (0,7)$ and slope $− 3 2 − 3 2$ 73. A y-intercept of $(0,3) (0,3)$ and slope $2 5 2 5$ 74. Passing through the points $(–6,–2) (–6,–2)$ and $(6,–6) (6,–6)$ 75. Passing through the points $(–3,–4) (–3,–4)$ and $(3,0) (3,0)$ For the following exercises, sketch the graph of each equation. 76. $f( x )=−2x−1 f( x )=−2x−1$ 77. $f( x )=−3x+2 f( x )=−3x+2$ 78. $f( x )= 1 3 x+2 f( x )= 1 3 x+2$ 79. $f( x )= 2 3 x−3 f( x )= 2 3 x−3$ 80. $f( t )=3+2t f( t )=3+2t$ 81. $p( t )=−2+3t p( t )=−2+3t$ 82. $x=3 x=3$ 83. $x=−2 x=−2$ 84. $r( x )=4 r( x )=4$ For the following exercises, write the equation of the line shown in the graph. 85. 86. 87. 88. #### 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. 89.  $x x$ 0 5 10 15 $g( x ) g( x )$ 5 –10 –25 –40 90.  $xx$ 0 5 10 15 $h( x ) h( x )$ 5 30 105 230 91.  $x x$ 0 5 10 15 $f( x ) f( x )$ –5 20 45 70 92.  $x x$ 5 10 20 25 $k( x ) k( x )$ 13 28 58 73 93.  $x x$ 0 2 4 6 $g( x ) g( x )$ 6 –19 –44 –69 94.  $x x$ 2 4 8 10 $h( x ) h( x )$ 13 23 43 53 95.  $x x$ 2 4 6 8 $f( x ) f( x )$ –4 16 36 56 96.  $x x$ 0 2 6 8 $k( x ) k( x )$ 6 31 106 231 #### Technology For the following exercises, use a calculator or graphing technology to complete the task. 97. If $f f$ is a linear function, $f(0.1)=11.5f(0.1)=11.5$ , and $f(0.4)=–5.9 f(0.4)=–5.9$ , find an equation for the function. 98. Graph the function $f f$ on a domain of $[–10,10]:f(x)=0.02x−0.01. [–10,10]:f(x)=0.02x−0.01.$ Enter the function in a graphing utility. For the viewing window, set the minimum value of $x x$ to be $−10 −10$ and the maximum value of $x x$ to be $10 10$ . 99. Graph the function $f f$ on a domain of $[–10,10]:fx)=2,500x+4,000 [–10,10]:fx)=2,500x+4,000$ 100. Table 3 shows the input, $w, w,$ and output, $k, k,$ for a linear function $k. k.$ 1. Fill in the missing values of the table. 2. Write the linear function $k, k,$ round to 3 decimal places.  w –10 5.5 67.5 b k 30 –26 a –44 Table 3 101. Table 4 shows the input, $p, p,$ and output, $q, q,$ for a linear function $q. q.$ 1. Fill in the missing values of the table. 2. Write the linear function $k. k.$  p 0.5 0.8 12 b q 400 700 a 1,000,000 Table 4 102. Graph the linear function $f f$ on a domain of $[ −10,10 ] [ −10,10 ]$ for the function whose slope is $1 8 1 8$ and y-intercept is $31 16 . 31 16 .$ Label the points for the input values of $−10 −10$ and $10. 10.$ 103. Graph the linear function $f f$ on a domain of $[ −0.1,0.1 ] [ −0.1,0.1 ]$ for the function whose slope is 75 and y-intercept is $−22.5. −22.5.$ Label the points for the input values of $−0.1 −0.1$ and $0.1. 0.1.$ 104. Graph the linear function $f f$ where $f( x )=ax+b f( x )=ax+b$ on the same set of axes on a domain of $[ −4,4 ] [ −4,4 ]$ for the following values of $a a$ and $b. b.$ 1. $a=2;b=3 a=2;b=3$ 2. $a=2;b=4 a=2;b=4$ 3. $a=2;b=–4 a=2;b=–4$ 4. $a=2;b=–5 a=2;b=–5$ #### Extensions 105. Find the value of $x x$ if a linear function goes through the following points and has the following slope: $(x,2),(−4,6),m=3 (x,2),(−4,6),m=3$ 106. Find the value of y if a linear function goes through the following points and has the following slope: $(10,y),(25,100),m=−5 (10,y),(25,100),m=−5$ 107. Find the equation of the line that passes through the following points: $( a,b ) ( a,b )$ and $( a,b+1 ) ( a,b+1 )$ 108. Find the equation of the line that passes through the following points: $(2a,b) (2a,b)$ and $(a,b+1) (a,b+1)$ 109. Find the equation of the line that passes through the following points: $(a,0) (a,0)$ and $(c,d) (c,d)$ 110. Find the equation of the line parallel to the line $g( x )=−0.01x+2.01 g( x )=−0.01x+2.01$ through the point $(1,2). (1,2).$ 111. Find the equation of the line perpendicular to the line $g( x )=−0.01x+2.01 g( x )=−0.01x+2.01$ through the point $(1,2). (1,2).$ For the following exercises, use the functions 112. Find the point of intersection of the lines $f f$ and $g. g.$ 113. Where is $f( x ) f( x )$ greater than $g( x )? g( x )?$ Where is $g( x ) g( x )$ greater than $f( x )? f( x )?$ #### Real-World Applications 114. At noon, a barista notices that she has$20 in her tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves $n n$ more customers during her shift? 115. 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? 116. A clothing business finds there is a linear relationship between the number of shirts, $n, n,$ it can sell and the price, $p, p,$ it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, 30,$ while 3,000 shirts can be sold at a price of$22. Find a linear equation in the form $p(n)=mn+b p(n)=mn+b$ that gives the price $p p$ they can charge for $n n$ shirts.

117.

A phone company charges for service according to the formula: $C(n)=24+0.1n, C(n)=24+0.1n,$ where $n n$ is the number of minutes talked, and $C(n) C(n)$ is the monthly charge, in dollars. Find and interpret the rate of change and initial value.

118.

A farmer finds there is a linear relationship between the number of bean stalks, $n, n,$ she plants and the yield, $y, y,$ 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 $y=mn+b y=mn+b$ that gives the yield when $n n$ stalks are planted.

119.

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.

120.

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, $P(t), P(t),$ for the population $t t$ years after 2003.

121.

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

1. As of 1990, average annual income was $23,286. 2. In the ten-year period from 1990–1999, average annual income increased by a total of$1,054.
3. Each year in the decade of the 1990s, average annual income increased by $1,054. 4. Average annual income rose to a level of$23,286 by the end of 1999.
122.

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 $C, C,$ the Celsius temperature, $F( C ). F( C ).$

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

### Footnotes

• 2http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm
• 3http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/