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) so this is the y-intercept.

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). 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

Substituting the slope and y-intercept into the slope-intercept form of a line gives

Example 4: Matching Linear Functions to Their Graphs

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

1. [latex]f\left(x\right)=2x+3[/latex]
2. [latex]g\left(x\right)=2x - 3[/latex]
3. [latex]h\left(x\right)=-2x+3[/latex]
4. [latex]j\left(x\right)=\frac{1}{2}x+3[/latex]

 

Figure 9

Show Solution

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 g 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) so f must be represented by line I.
2. This function also has a slope of 2, but a y-intercept of –3. It must pass through the point (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 [latex]\frac{1}{2}[/latex] 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), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II.

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

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) equal to zero and solve for the value of x. For example, consider the function shown.

Set the function equal to 0 and solve for x.

The graph of the function crosses the x-axis at the point (2, 0).

Figure 11

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 = 0 in the equation [latex]f(x)=mx+b[/latex], the equation simplifies to [latex]f(x)=b[/latex]. In other words, the value of the function is a constant. This graph represents the function [latex]f(x)=2[/latex].

Figure 14

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.

Notice that a vertical line, such as the one in Figure 15, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2.

Example 6: Writing the Equation of a Horizontal Line

Write the equation of the line graphed in Figure 16.

Figure 16

Show Solution

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

Example 7: Writing the Equation of a Vertical Line

Write the equation of the line graphed in Figure 17.

Figure 17

Show Solution

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

Determining Whether Lines are Parallel or Perpendicular

The two lines in Figure 18 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.

Figure 18 Parallel lines.

Figure 19

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 20 are perpendicular.

Figure 20. 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 [latex]{m}_{1}\text{ and }{m}_{2}[/latex] are negative reciprocals of one another, they can be multiplied together to yield [latex]-1[/latex].

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is [latex]\frac{1}{8}[/latex], and the reciprocal of [latex]\frac{1}{8}[/latex] 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 perpendicular. 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.

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.

[latex]\begin{align}&f\left(x\right)=2x+3 &&h\left(x\right)=-2x+2 \\ &g\left(x\right)=\frac{1}{2}x - 4 && j\left(x\right)=2x - 6 \end{align}[/latex]

Show Solution

Parallel lines have the same slope. Because the functions [latex]f\left(x\right)=2x+3[/latex] and [latex]j\left(x\right)=2x - 6[/latex] each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and [latex]\frac{1}{2}[/latex] are negative reciprocals, the equations, [latex]g\left(x\right)=\frac{1}{2}x - 4[/latex] and [latex]h\left(x\right)=-2x+2[/latex] represent perpendicular lines.

Analysis of the Solution

A graph of the lines is shown in Figure 21.

Figure 21. The graph shows that the lines [latex]f\left(x\right)=2x+3[/latex] and [latex]j\left(x\right)=2x - 6[/latex] are parallel, and the lines [latex]g\left(x\right)=\frac{1}{2}x - 4[/latex] and [latex]h\left(x\right)=-2x+2[/latex] 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.

[latex]f\left(x\right)=3x+1[/latex]

We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to f(x). So the lines formed by all of the following functions will be parallel to f(x).

[latex]\begin{align}&g\left(x\right)=3x+6 \\ &h\left(x\right)=3x+1 \\ &p\left(x\right)=3x+\frac{2}{3} \end{align}[/latex]

Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for 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.

[latex]\begin{gathered}y-{y}_{1}=m\left(x-{x}_{1}\right) \\ y - 7=3\left(x - 1\right) \\ y - 7=3x - 3 \\ y=3x+4 \end{gathered}[/latex]

So [latex]g(x)=3x+4[/latex] is parallel to [latex]f(x)=3x+1[/latex] and passes through the point (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 [latex]f\left(x\right)=3x+6[/latex] that passes through the point (3,0).

Show Solution

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

[latex]\begin{gathered}g\left(x\right)=3x+b \\ 0=3\left(3\right)+b \\ b=-9 \end{gathered}[/latex]

The line parallel to f(x) that passes through (3, 0) is [latex]g\left(x\right)=3x - 9[/latex].

Analysis of the Solution

We can confirm that the two lines are parallel by graphing them. Figure 22 shows that the two lines will never intersect.

Figure 22

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:

[latex]f\left(x\right)=2x+4[/latex]

The slope of the line is 2, and its negative reciprocal is [latex]-\frac{1}{2}[/latex]. Any function with a slope of [latex]-\frac{1}{2}[/latex] will be perpendicular to f(x). So the lines formed by all of the following functions will be perpendicular to f(x).

[latex]\begin{align}&g\left(x\right)=-\frac{1}{2}x+4 \\ &h\left(x\right)=-\frac{1}{2}x+2 \\ &p\left(x\right)=-\frac{1}{2}x-\frac{1}{2} \end{align}[/latex]

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) and passes through the point (4, 0). We already know that the slope is [latex]-\frac{1}{2}[/latex]. 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.

[latex]\begin{gathered}g\left(x\right)=mx+b \\ 0=-\frac{1}{2}\left(4\right)+b \\ 0=-2+b \\ 2=b \\ b=2 \end{gathered}[/latex]

The equation for the function with a slope of [latex]-\frac{1}{2}[/latex] and a y-intercept of 2 is

[latex]g\left(x\right)=-\frac{1}{2}x+2[/latex].

So [latex]g\left(x\right)=-\frac{1}{2}x+2[/latex] is perpendicular to [latex]f\left(x\right)=2x+4[/latex] and passes through the point (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 and y from the coordinate pair provided into [latex]g\left(x\right)=mx+b[/latex].
4. Solve for 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 [latex]f\left(x\right)=3x+3[/latex] that passes through the point (3,0).

Show Solution

The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\frac{1}{3}[/latex]. Using this slope and the given point, we can find the equation for the line.

[latex]\begin{gathered}g\left(x\right)=-\frac{1}{3}x+b \\ 0=-\frac{1}{3}\left(3\right)+b \\ 1=b \\ b=1 \end{gathered}[/latex]

The line perpendicular to f(x) that passes through (3, 0) is [latex]g\left(x\right)=-\frac{1}{3}x+1[/latex].

Analysis of the Solution

A graph of the two lines is shown in Figure 23.

Figure 23

Try It

Given the function [latex]h\left(x\right)=2x - 4[/latex], write an equation for the line passing through (0, 0) that is

a. parallel to h(x)
b. perpendicular to h(x)

Show Solution

a. [latex]f\left(x\right)=2x[/latex]
b. [latex]g\left(x\right)=-\frac{1}{2}x[/latex]

Try It

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) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).

Show Solution

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

[latex]\begin{align}{m}_{1}&=\frac{5 - 6}{4-\left(-2\right)} \\ &=\frac{-1}{6} \\ &=-\frac{1}{6} \end{align}[/latex]

Find the negative reciprocal of the slope.

[latex]\begin{align}{m}_{2}&=\frac{-1}{-\frac{1}{6}} \\ &=-1\left(-\frac{6}{1}\right) \\ &=6 \end{align}[/latex]

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

[latex]\begin{gathered}g\left(x\right)=6x+b \\ 5=6\left(4\right)+b \\ 5=24+b \\ -19=b \\ b=-19 \end{gathered}[/latex]

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

[latex]y=6x - 19[/latex]

Try It

A line passes through the points, (–2, –15) and (2, –3). Find the equation of a perpendicular line that passes through the point, (6, 4).

Show Solution

[latex]y=-\frac{1}{3}x+6[/latex]

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 [latex]f\left(x\right)=g\left(x\right)[/latex]. 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 [latex]h\left(t\right)=3t - 4[/latex] and [latex]j\left(t\right)=5-t[/latex].

Show Solution

Set [latex]h\left(t\right)=j\left(t\right)[/latex].

[latex]\begin{align}3t - 4&=5-t \\ 4t&=9 \\ t&=\frac{9}{4} \end{align}[/latex]

This tells us the lines intersect when the input is [latex]\frac{9}{4}[/latex].

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

[latex]\begin{align}j\left(\frac{9}{4}\right)&=5-\frac{9}{4} \\ &=\frac{11}{4} \end{align}[/latex]

These lines intersect at the point [latex]\left(\frac{9}{4},\frac{11}{4}\right)[/latex].

Analysis of the Solution

Looking at Figure 24, this result seems reasonable.

Figure 24

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

Using the graph in the Analysis of the Solution for Example 12, identify the following for the function [latex]j\left(t\right)[/latex]:

a. y-intercept

b. x-intercept(s)

c. slope

d. Is [latex]j\left(t\right)[/latex] parallel or perpendicular to [latex]h\left(t\right)[/latex] (or neither)?

e. Is [latex]j\left(t\right)[/latex] an increasing or decreasing function (or neither)?

f. Write a transformation description for [latex]j\left(t\right)[/latex] from the identity toolkit function [latex]f\left(x\right)=x[/latex].

Show Solution

a. [latex]\left(0,5\right)[/latex]

b. [latex]\left(5,\text{ 0}\right)[/latex]

c. Slope -1

d. Neither parallel nor perpendicular

e. Decreasing function

f. Given the identity function, perform a vertical flip (over the t-axis) and shift up 5 units.

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

Show Solution

1. he cost function in the sum of the fixed cost, $125,000, and the variable cost, $120 per helmet.
   [latex]C\left(x\right)=120x+250,000[/latex]
2. The revenue function is the total revenue from the sale of [latex]x[/latex] helmets, [latex]R\left(x\right)=140x[/latex].
3. The break-even point is the point of intersection of the graph of the cost and revenue functions. To find the x-coordinate of the coordinate pair of the point of intersection, set the two equations equal, and solve for x.

[latex]\begin{align}C\left(x\right)&=R\left(x\right) \\ 250,000+120x&=140x \\ 250,000&=20x \\ 12,500&=x \\ x&=12,500 \end{align}[/latex]

To find [latex]y[/latex], evaluate either the revenue or the cost function at 12,500.

[latex]\begin{align}R\left(20\right)&=140\left(12,500\right) \\ &=$1,750,000 \end{align}[/latex]

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

Analysis of the Solution

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

Figure 25

Key Concepts

• Linear functions may be graphed by plotting points or by using the y-intercept and slope.
• Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections.
• The y-intercept and slope of a line may be used to write the equation of a line.
• The x-intercept is the point at which the graph of a linear function crosses the x-axis.
• Horizontal lines are written in the form, f(x) = b.
• Vertical lines are written in the form, x = b.
• Parallel lines have the same slope.
• Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.
• A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x– and y-values of the given point into the equation, [latex]f(x)=mx+b[/latex], and using the b that results. Similarly, the point-slope form of an equation can also be used.
• A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.
• A system of linear equations may be solved setting the two equations equal to one another and solving for x. The y-value may be found by evaluating either one of the original equations using this x-value.
• A system of linear equations may also be solved by finding the point of intersection on a graph.

Glossary

horizontal line

a line defined by [latex]f(x)=b[/latex], where b is a real number. The slope of a horizontal line is 0.

parallel lines

two or more lines with the same slope

perpendicular lines

two lines that intersect at right angles and have slopes that are negative reciprocals of each other

vertical line

a line defined by x = a, where a is a real number. The slope of a vertical line is undefined.

x-intercept

the point on the graph of a linear function when the output value is 0; the point at which the graph crosses the horizontal axis