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2.3: Graphs of Linear Functions

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    113995
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    Learning Objectives

    In this section, you will:

    • Graph linear functions.
    • Write the equation for a linear function from the graph of a line.
    • Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
    • Write the equation of a line parallel or perpendicular to a given line.
    • Solve a system of linear equations.

    Two competing telephone companies offer different payment plans. The two plans charge the same rate per long distance minute, but charge a different monthly flat fee. A consumer wants to determine whether the two plans will ever cost the same amount for a given number of long distance minutes used. The total cost of each payment plan can be represented by a linear function. To solve the problem, we will need to compare the functions. In this section, we will consider methods of comparing functions using graphs.

    Graphing Linear Functions

    In Linear Functions, we saw that that the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their characteristics.

    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 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 1

    Graphing by Plotting Points

    Graph f(x)=23x+5f(x)=23x+5 by plotting points.

    Answer

    Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6.

    Evaluate the function at each input value, and use the output value to identify coordinate pairs.

    x=0 f(0)= 2 3 (0)+5=5( 0,5 ) x=3 f(3)= 2 3 (3)+5=3( 3,3 ) x=6 f(6)= 2 3 (6)+5=1( 6,1 ) x=0 f(0)= 2 3 (0)+5=5( 0,5 ) x=3 f(3)= 2 3 (3)+5=3( 3,3 ) x=6 f(6)= 2 3 (6)+5=1( 6,1 )

    Plot the coordinate pairs and draw a line through the points. Figure 1 represents the graph of the function f(x)=23x+5.f(x)=23x+5.

    24f1218b070a0504b55038c27f2fc9790304a136
    Figure 1 The graph of the linear function f ( x ) = 2 3 x + 5. f ( x ) = 2 3 x + 5.

    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 #1

    Graph f(x)=34x+6f(x)=34x+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=0x=0 in the equation.

    The other characteristic of the linear function is its slope m,m, which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a function is equal to the ratio of the change in outputs to the change in inputs. Another way to think about the slope is by dividing the vertical difference, or rise, by the horizontal difference, or run. We encountered both the y-intercept and the slope in Linear Functions.

    Let’s consider the following function.

    f(x)=12x+1f(x)=12x+1

    The slope is 12.12. 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 rise over run, m=riserun.m=riserun. From our example, we have m=12,m=12, 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 2.

    fa8d440278a4d73105cd08756daceb00bac43939
    Figure 2

    Graphical Interpretation of a Linear Function

    In the equation f(x)=mx+bf(x)=mx+b

    • bb is the y-intercept of the graph and indicates the point (0,b)(0,b) at which the graph crosses the y-axis.
    • mm 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:

    m= change in output (rise) change in input (run) = Δy Δx = y 2 y 1 x 2 x 1 m= change in output (rise) change in input (run) = Δy Δx = y 2 y 1 x 2 x 1

    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 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 riserunriserun to determine at least two more points on the line.
    5. Sketch the line that passes through the points.

    Example 2

    Graphing by Using the y-intercept and Slope

    Graph f(x)=23x+5f(x)=23x+5 using the y-intercept and slope.

    Answer

    Evaluate the function at x=0x=0 to find the y-intercept. The output value when x=0x=0 is 5, so the graph will cross the y-axis at (0,5).(0,5).

    According to the equation for the function, the slope of the line is 23.23. This tells us that for each vertical decrease in the “rise” of 22 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 3. From the initial value (0,5)(0,5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points.

    06eec60511986086ec1b534fd3082df0d9f7be5d
    Figure 3

    Analysis

    The graph slants downward from left to right, which means it has a negative slope as expected.

    Try It #2

    Find a point on the graph we drew in Example 2 that has a negative x-value.

    Graphing a Function Using Transformations

    Another option for graphing is to use transformations of the identity function f(x)=xf(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 mm is acting as the vertical stretch or compression of the identity function. When mm is negative, there is also a vertical reflection of the graph. Notice in Figure 4 that multiplying the equation of f(x)=xf(x)=x by mm stretches the graph of ff by a factor of mm units if m>1m>1 and compresses the graph of ff by a factor of mm units if 0<m<1.0<m<1. This means the larger the absolute value of m,m, the steeper the slope.

    c022489323343d3f67bbc9b025766d0cc7fbe1b2
    Figure 4 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 bb acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in Figure 5 that adding a value of bb to the equation of f(x)=xf(x)=x shifts the graph of ff a total of bb units up if bb is positive and |b||b| units down if bb is negative.

    0120bf1497785e3d10342ceca2ddf50bf6a0d285
    Figure 5 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 bb units.

    Example 3

    Graphing by Using Transformations

    Graph f(x)=12x3f(x)=12x3 using transformations.

    Answer

    The equation for the function shows that m=12m=12 so the identity function is vertically compressed by 12.12. The equation for the function also shows that b=−3b=−3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 6.

    d5621b67f6db03da1c64ec9b78f685850687e2a2
    Figure 6 The function, y = x , y = x , compressed by a factor of 1 2 . 1 2 .

    Then show the vertical shift as in Figure 7.

    fa6330d94e6f7cd4b4fcc0bd7d135a06f1c34dd2
    Figure 7 The function y = 1 2 x , y = 1 2 x , shifted down 3 units.

    Try It #3

    Graph f(x)=4+2x,f(x)=4+2x, using transformations.

    Q&A

    In Example 3, 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 =13 =2 f(2)= 1 2 (2)3 =13 =2

    Writing the Equation for a Function from the Graph of a Line

    Recall that in Linear Functions, 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 8. 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.

    20978dd1c42fbfaed1ae1cf037172664543b42fb
    Figure 8

    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

    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 4

    Matching Linear Functions to Their Graphs

    Match each equation of the linear functions with one of the lines in Figure 9.

    1. f(x)=2x+3f(x)=2x+3
    2. g(x)=2x3g(x)=2x3
    3. h(x)=2x+3h(x)=2x+3
    4. j(x)=12x+3j(x)=12x+3
    f7b53aade291d174c12865880e32db6f0bf8bb5d
    Figure 9
    Answer

    Analyze the information for each function.

    1. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function gg has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through ( 0, 3)( 0, 3) so ff must be represented by Line I.
    2. This function also has a slope of 2, but a y-intercept of 3.3. It must pass through the point (0,3)(0,3) and slant upward from left to right. It must be represented by Line III.
    3. This function has a slope of –2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
    4. This function has a slope of 1212 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3),(0, 3), but the slope of jj is less than the slope of ff so the line for jj must be flatter. This function is represented by Line II.

    Now we can re-label the lines as in Figure 10.

    909fbcd4ca1913a9ca3982b1f6378884c83b8fc8
    Figure 10

    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. 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)=3x6f(x)=3x6

    Set the function equal to 0 and solve for x.x.

    0=3x6 6=3x 2=x x=2 0=3x6 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 cc is a nonzero real number are the only examples of linear functions with no x-intercept. For example, y=5y=5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 11.

    Graph of y = 5.
    Figure 11

    x-intercept

    The x-intercept of the function is value of xx when f(x)=0.f(x)=0. It can be solved by the equation 0=mx+b.0=mx+b.

    Example 5

    Finding an x-intercept

    Find the x-intercept of f(x)=12x3.f(x)=12x3.

    Answer

    Set the function equal to zero to solve for x.x.

    0= 1 2 x3 3= 1 2 x 6=x x=6 0= 1 2 x3 3= 1 2 x 6=x x=6

    The graph crosses the x-axis at the point (6, 0).(6, 0).

    Analysis

    A graph of the function is shown in Figure 12. We can see that the x-intercept is (6, 0)(6, 0) as we expected.

    8504d6a1e9bda608b826e94f0b0ebbcd62e6f523
    Figure 12 The graph of the linear function f ( x ) = 1 2 x 3. f ( x ) = 1 2 x 3.

    Try It #4

    Find the x-intercept of f(x)=14x4.f(x)=14x4.

    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 13, 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=0m=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.

    ef1eca76a5381668fc88461826818c2a424abc92
    Figure 13 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.

    7f86347d3fbfab18c5d60f08b188e16a3be7c53d

    Notice that a vertical line, such as the one in Figure 14, 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.

    1f5db247564ebd77f64450ef6f56236c697c9845
    Figure 14 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 6

    Writing the Equation of a Horizontal Line

    Write the equation of the line graphed in Figure 15.

    Graph of x = 7.
    Figure 15
    Answer

    For any x-value, the y-value is 4,4, so the equation is y=4.y=4.

    Example 7

    Writing the Equation of a Vertical Line

    Write the equation of the line graphed in Figure 16.

    Graph of two functions where the baby blue line is y = -2/3x + 7, and the blue line is y = -x + 1.
    Figure 16
    Answer

    The constant x-value is 7,7, so the equation is x=7.x=7.

    Determining Whether Lines are Parallel or Perpendicular

    The two lines in Figure 17 are parallel lines: they will never intersect. Notice that 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 y-intercept of the other, they would become the same line.

    Graph of two functions where the blue line is y = -2/3x + 1, and the baby blue line is y = -2/3x +7. Notice that they are parallel lines.
    Figure 17 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.

    f(x)=2x+6 f(x)=2x4 }parallel f(x)=3x+2 f(x)=2x+2 }not parallel f(x)=2x+6 f(x)=2x4 }parallel f(x)=3x+2 f(x)=2x+2 }not parallel

    Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 18 are perpendicular.

    Graph of two functions where the blue line is perpendicular to the orange line.
    Figure 18 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. So, if m1m1 and m2m2 are negative reciprocals of one another, they can be multiplied together to yield –1.

    m1m2=1m1m2=1

    To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is 18,18, and the reciprocal of 1818 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.

    f(x)= 1 4 x+2 negative reciprocal of 1 4 is −4 f(x)=4x+3 negative reciprocal of4 is 1 4 f(x)= 1 4 x+2 negative reciprocal of 1 4 is −4 f(x)=4x+3 negative reciprocal of4 is 1 4

    The product of the slopes is –1.

    4(14)=14(14)=1

    Parallel and Perpendicular Lines

    Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

    f(x)=m1x+b1 and g(x)=m2x+b2 are parallel if m1=m2.f(x)=m1x+b1 and g(x)=m2x+b2 are parallel if m1=m2.

    If and only if b1=b2b1=b2 and m1=m2,m1=m2, we say the lines coincide. Coincident lines are the same line.

    Two lines are perpendicular lines if they intersect at right angles.

    f(x)= m 1 x+ b 1 and g(x)= m 2 x+ b 2 are perpendicular if m 1 m 2 =1, and so m 2 = 1 m 1 . f(x)= m 1 x+ b 1 and g(x)= m 2 x+ b 2 are perpendicular if m 1 m 2 =1, and so m 2 = 1 m 1 .

    Example 8

    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 x4 j(x)=2x6 f(x)=2x+3 h(x)=2x+2 g(x)= 1 2 x4 j(x)=2x6

    Answer

    Parallel lines have the same slope. Because the functions f(x)=2x+3f(x)=2x+3 and j(x)=2x6 j(x)=2x6 each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and 1212 are negative reciprocals, the equations, g(x)=12x4g(x)=12x4 and h(x)=2x+2h(x)=2x+2 represent perpendicular lines.

    Analysis

    A graph of the lines is shown in Figure 19.

    Graph of four functions where the blue line is h(x) = -2x + 2, the orange line is f(x) = 2x + 3, the green line is j(x) = 2x - 6, and the red line is g(x) = 1/2x - 4.
    Figure 19

    The graph shows that the lines f(x)=2x+3f(x)=2x+3 and j(x)=2x6j(x)=2x6 are parallel, and the lines g(x)=12x4g(x)=12x4 and h(x)=2x+2h(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 following equation.

    f(x)=3x+1f(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 ff and passes through the point (1, 7).(1, 7). We already know that the slope is 3. We just need to determine which value for bb 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 ) y7=3(x1) y7=3x3 y=3x+4 y y 1 =m(x x 1 ) y7=3(x1) y7=3x3 y=3x+4

    So g(x)=3x+4g(x)=3x+4 is parallel to f(x)=3x+1f(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 9

    Finding a Line Parallel to a Given Line

    Find a line parallel to the graph of f(x)=3x+6f(x)=3x+6 that passes through the point (3, 0).(3, 0).

    Answer

    The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m=3,m=3, x=3,x=3, and f(x)=0f(x)=0 into the slope-intercept form to find the y-intercept.

    g(x)=3x+b 0=3(3)+b b=9 g(x)=3x+b 0=3(3)+b b=9

    The line parallel to f(x)f(x) that passes through (3, 0)(3, 0) is g(x)=3x9.g(x)=3x9.

    Analysis

    We can confirm that the two lines are parallel by graphing them. Figure 20 shows that the two lines will never intersect.

    Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.
    Figure 20

    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 following function:

    f(x)=2x+4f(x)=2x+4

    The slope of the line is 2, and its negative reciprocal is 12.12. Any function with a slope of 1212 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 12.12. 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 1212 and a y-intercept of 2 is

    g(x)=12x+2.g(x)=12x+2.

    So g(x)=12x+2g(x)=12x+2 is perpendicular to f(x)=2x+4f(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 xx and yy from the coordinate pair provided into g(x)=mx+b.g(x)=mx+b.
    4. Solve for b.b.
    5. Write the equation for the line.

    Example 10

    Finding the Equation of a Perpendicular Line

    Find the equation of a line perpendicular to f(x)=3x+3f(x)=3x+3 that passes through the point (3, 0).(3, 0).

    Answer

    The original line has slope m=3,m=3, so the slope of the perpendicular line will be its negative reciprocal, or 13.13. Using this slope and the given point, we can find the equation for the line.

    g(x)= 1 3 x+b 0= 1 3 (3)+b 1=b b=1 g(x)= 1 3 x+b 0= 1 3 (3)+b 1=b b=1

    The line perpendicular to f(x)f(x) that passes through (3, 0)(3, 0) is g(x)=13x+1.g(x)=13x+1.

    Analysis

    A graph of the two lines is shown in Figure 21 below.

    Graph of two functions where the blue line is g(x) = -1/3x + 1, and the orange line is f(x) = 3x + 6.
    Figure 21

    Try It #5

    Given the function h(x)=2x4,h(x)=2x4, 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 11

    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).

    Answer

    From the two points of the given line, we can calculate the slope of that line.

    m 1 = 56 4(2) = 1 6 = 1 6 m 1 = 56 4(2) = 1 6 = 1 6

    Find the negative reciprocal of the slope.

    m 2 = 1 1 6 =1( 6 1 ) =6 m 2 = 1 1 6 =1( 6 1 ) =6

    We can then solve for the y-intercept of the line passing through the point (4,5).(4,5).

    g(x)=6x+b 5=6(4)+b 5=24+b 19=b b=−19 g(x)=6x+b 5=6(4)+b 5=24+b 19=b b=−19

    The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4,5)(4,5) is

    y=6x19y=6x19

    Try It #6

    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).

    Solving a System of Linear Equations Using a Graph

    A system of linear equations includes two or more linear equations. The graphs of two lines will intersect at a single point if they are not parallel. Two parallel lines can also intersect if they are coincident, which means they are the same line and they intersect at every point. For two lines that are not parallel, the single point of intersection will satisfy both equations and therefore represent the solution to the system.

    To find this point when the equations are given as functions, we can solve for an input value so that f(x)=g(x).f(x)=g(x). In other words, we can set the formulas for the lines equal to one another, and solve for the input that satisfies the equation.

    Example 12

    Finding a Point of Intersection Algebraically

    Find the point of intersection of the lines h(t)=3t4h(t)=3t4 and j(t)=5t.j(t)=5t.

    Answer

    Set h(t)=j(t).h(t)=j(t).

    3t4=5t 4t=9 t= 9 4 3t4=5t 4t=9 t= 9 4

    This tells us the lines intersect when the input is 94.94.

    We can then find the output value of the intersection point by evaluating either function at this input.

    j( 9 4 )=5 9 4 = 11 4 j( 9 4 )=5 9 4 = 11 4

    These lines intersect at the point (94,114).(94,114).

    Analysis

    Looking at Figure 22, this result seems reasonable.

    Graph of two functions h(t) = 3t - 4 and j(t) = t +5 and their intersection at (9/4, 11/4).
    Figure 22

    Q&A

    If we were asked to find the point of intersection of two distinct parallel lines, should something in the solution process alert us to the fact that there are no solutions?

    Yes. After setting the two equations equal to one another, the result would be the contradiction “0 = non-zero real number”.

    Try It #7

    Look at the graph in Figure 22 and identify the following for the function j(t):j(t):

    1. y-intercept
    2. x-intercept(s)
    3. slope
    4. Is j(t)j(t) parallel or perpendicular to h(t)h(t) (or neither)?
    5. Is j(t)j(t) an increasing or decreasing function (or neither)?
    6. Write a transformation description for j(t)j(t) from the identity toolkit function f(x)=x.f(x)=x.

    Example 13

    Finding a Break-Even Point

    A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120 to produce, and sells for $140.

    1. Find the cost function, C,C, to produce xx helmets, in dollars.
    2. Find the revenue function, R,R, from the sales of xx helmets, in dollars.
    3. Find the break-even point, the point of intersection of the two graphs CC and R.R.
    Answer

    1. The cost function in the sum of the fixed cost, $125,000, and the variable cost, $120 per helmet.

      C(x)=120x+250,000C(x)=120x+250,000

    2. The revenue function is the total revenue from the sale of xx helmets, R(x)=140x.R(x)=140x.
    3. C(x)=R(x) 250,000+120x=140x 250,000=20x 12,500=x x=12,500 C(x)=R(x) 250,000+120x=140x 250,000=20x 12,500=x x=12,500

      R(x)=140(12,500) =$1,750,000 R(x)=140(12,500) =$1,750,000

    The break-even point is (12,500,1,750,000). (12,500,1,750,000).

    Analysis

    This means if the company sells 12,500 helmets, they break even; both the sales and cost incurred equaled 1.75 million dollars. See Figure 23

    Graph of the two functions, C(x) and R(x) where it shows that below (12500, 1750000) the company loses money and above that point the company makes a profit.
    Figure 23

    Media

    Access these online resources for additional instruction and practice with graphs of linear functions.

    2.2 Section Exercises

    Verbal

    1.

    If the graphs of two linear functions are parallel, describe the relationship between the slopes and the y-intercepts.

    2.

    If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y-intercepts.

    3.

    If a horizontal line has the equation f(x)=af(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.

    4.

    Explain how to find a line parallel to a linear function that passes through a given point.

    5.

    Explain how to find a line perpendicular to a linear function that passes through a given point.

    Algebraic

    For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular:

    6.

    4x7y=10 7x+4y=1 4x7y=10 7x+4y=1

    7.

    3y+x=12 y=8x+1 3y+x=12 y=8x+1

    8.

    3y+4x=12 6y=8x+1 3y+4x=12 6y=8x+1

    9.

    6x9y=10 3x+2y=1 6x9y=10 3x+2y=1

    10.

    y= 2 3 x+1 3x+2y=1 y= 2 3 x+1 3x+2y=1

    11.

    y= 3 4 x+1 3x+4y=1 y= 3 4 x+1 3x+4y=1

    For the following exercises, find the x- and y-intercepts of each equation

    12.

    f( x )=x+2 f( x )=x+2

    13.

    g(x)=2x+4g(x)=2x+4

    14.

    h( x )=3x5 h( x )=3x5

    15.

    k( x )=5x+1 k( x )=5x+1

    16.

    2x+5y=20 2x+5y=20

    17.

    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?

    18.
    • 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)
    19.
    • 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)
    20.
    • 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)
    21.
    • 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)
    22.
    • Line 1: Passes through (0,5)(0,5) and (3,3)(3,3)
    • Line 2: Passes through (1,−5)(1,−5) and (3,−2)(3,−2)
    23.
    • 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)
    24.

    Write an equation for a line parallel to f(x)=5x3f(x)=5x3 and passing through the point (2,12).(2,12).

    25.

    Write an equation for a line parallel to g(x)=3x1g(x)=3x1 and passing through the point (4,9).(4,9).

    26.

    Write an equation for a line perpendicular to h(t)=2t+4h(t)=2t+4 and passing through the point (-4,1).(-4,1).

    27.

    Write an equation for a line perpendicular to p(t)=3t+4p(t)=3t+4 and passing through the point (3,1).(3,1).

    28.

    Find the point at which the line f(x)=2x1f(x)=2x1 intersects the line g(x)=x.g(x)=x.

    29.

    Find the point at which the line f(x)=2x+5f(x)=2x+5 intersects the line g(x)=3x5.g(x)=3x5.

    30.

    Use algebra to find the point at which the line f(x)=45x+27425f(x)=45x+27425 intersects the line h(x)=94x+7310.h(x)=94x+7310.

    31.

    Use algebra to find the point at which the line f(x)=74x+45760f(x)=74x+45760 intersects the line g(x)=43x+315.g(x)=43x+315.

    Graphical

    For the following exercises, match the given linear equation with its graph in Figure 24.

    a39e9ff4d84a7c0763a46ce587744ccb05e4d3c5
    Figure 24
    32.

    f( x )=x1 f( x )=x1

    33.

    f(x)=2x1f(x)=2x1

    34.

    f(x)=12x1f(x)=12x1

    35.

    f(x)=2f(x)=2

    36.

    f(x)=2+xf(x)=2+x

    37.

    f(x)=3x+2f(x)=3x+2

    For the following exercises, sketch a line with the given features.

    38.

    An x-intercept of (4, 0)(4, 0) and y-intercept of (0, –2)(0, –2)

    39.

    An x-intercept of (2, 0)(2, 0) and y-intercept of (0, 4)(0, 4)

    40.

    A y-intercept of (0, 7)(0, 7) and slope 3232

    41.

    A y-intercept of (0, 3)(0, 3) and slope 2525

    42.

    Passing through the points (6, –2)(6, –2) and (6, –6)(6, –6)

    43.

    Passing through the points (3, –4)(3, –4) and (3, 0) (3, 0)

    For the following exercises, sketch the graph of each equation.

    44.

    f(x)=2x1f(x)=2x1

    45.

    g(x)=3x+2g(x)=3x+2

    46.

    h(x)=13x+2h(x)=13x+2

    47.

    k(x)=23x3k(x)=23x3

    48.

    f( t )=3+2t f( t )=3+2t

    49.

    p(t)=2+3tp(t)=2+3t

    50.

    x=3x=3

    51.

    x=2 x=2

    52.

    r(x)=4r(x)=4

    53.

    q(x)=3q(x)=3

    54.

    4x=9y+364x=9y+36

    55.

    x3y4=1x3y4=1

    56.

    3x5y=153x5y=15

    57.

    3x=153x=15

    58.

    3y=123y=12

    59.

    If g(x)g(x) is the transformation of f(x)=xf(x)=x after a vertical compression by 3 4 , 3 4 , a shift right by 2, and a shift down by 4

    1. Write an equation for g(x).g(x).
    2. What is the slope of this line?
    3. Find the y-intercept of this line.
    60.

    If g(x)g(x) is the transformation of f(x)=xf(x)=x after a vertical compression by 13,13, a shift left by 1, and a shift up by 3

    1. Write an equation for g(x).g(x).
    2. What is the slope of this line?
    3. Find the y-intercept of this line.

    For the following exercises,, write the equation of the line shown in the graph.

    61.
    33c38da1736b288b7f138eecf0d7998aacacd314
    62.
    5871f0ae9e05bf660ad09386eaeaf534c6258481
    63.
    739ba6401964d68135f4f1bcb29d74066be1e991
    64.
    55091e58610407d19d57b76e71ed7d457b65cccc

    For the following exercises, find the point of intersection of each pair of lines if it exists. If it does not exist, indicate that there is no point of intersection.

    65.

    y= 3 4 x+1 3x+4y=12 y= 3 4 x+1 3x+4y=12

    66.

    2x3y=12 5y+x=30 2x3y=12 5y+x=30

    67.

    2x=y3y+4x=15 2x=y3y+4x=15

    68.

    x2y+2=3 xy=3 x2y+2=3 xy=3

    69.

    5x+3y=65 xy=5 5x+3y=65 xy=5

    Extensions

    70.

    Find the equation of the line parallel to the line g(x)=0.01x+2.01g(x)=0.01x+2.01 through the point (1, 2).(1, 2).

    71.

    Find the equation of the line perpendicular to the line g(x)=0.01x+2.01g(x)=0.01x+2.01 through the point (1, 2).(1, 2).

    For the following exercises, use the functions f(x)=0.1x+200 and g(x)=20x+0.1.f(x)=0.1x+200 and g(x)=20x+0.1.

    72.

    Find the point of intersection of the lines ff and g.g.

    73.

    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

    74.

    A car rental company offers two plans for renting a car.

    • Plan A: $30 per day and $0.18 per mile
    • Plan B: $50 per day with free unlimited mileage

    How many miles would you need to drive for plan B to save you money?

    75.

    A cell phone company offers two plans for minutes.

    • Plan A: $20 per month and $1 for every one hundred texts.
    • Plan B: $50 per month with free unlimited texts.

    How many texts would you need to send per month for plan B to save you money?

    76.

    A cell phone company offers two plans for minutes.

    • Plan A: $15 per month and $2 for every 300 texts.
    • Plan B: $25 per month and $0.50 for every 100 texts.

    How many texts would you need to send per month for plan B to save you money?


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