This section is a review of skills and concepts from your Beginning Algebra course. Feel free to skim the explanations and try the examples, or to work on the homework and come back for more help. If you find that you need more help understanding slope, please click on the link to learn more about slope.

Determine whether equations are linear or nonlinear

When you graph a linear equation, it forms a straight line on the coordinate plane. Linear equations have a constant rate of change which is called the slope. In contrast, nonlinear equations have graphs that do not form straight lines. The rate of change of a nonlinear function is not constant.

linear vs. non-linear functions

Linear	Non-linear
A function that can be written in the form [latex] f(x)=mx+b [/latex]	A function that CANNOT be written in the form [latex] f(x)=mx+b [/latex]
Each variable in the function has a maximum of degree 1.	At least one variable has a degree of 2 or more, or a variable appears in the denominator, or a variable is inside a square root or in an exponent (as a few examples)
The function when graphed is a straight line.	The function when graphed is not a straight line but is a curve.
Examples: [latex] f(x)=-3x [/latex] [latex] 7y=5x [/latex] [latex] f(x)=2x-9 [/latex] [latex] 3x-4y=12 [/latex] [latex] f(x)=\dfrac{1}{8}x [/latex]	Examples: [latex] f(x)=3x^2+4 [/latex] [latex] f(x)=2^x [/latex] [latex] f(x)=\dfrac{5}{x} [/latex] [latex] f(x)=4x^5-3x^4-1 [/latex] [latex] f(x)=3\sqrt{x}+2 [/latex]

nonlinear function	nonlinear function	linear function	nonlinear function
nonlinear function	linear function	nonlinear function	linear function

There are multiple ways to create a graph from a linear equation. One way is to create a table of values for [latex] x [/latex] and [latex] y [/latex] and then plot these ordered pairs on the coordinate plane. Two points are enough to determine a line. However, it is always a good idea to plot more than two points to avoid possible errors.

After graphing the points, draw a line through the points to show that all of the points are on the line. The arrows at each end of the graph indicate that the line continues endlessly in both directions. Every point on this line is a solution to the linear equation.

Example

Graph the linear equation [latex]y=2x-5[/latex].

Show Solution

Evaluate [latex]y=2x-5[/latex] for different values of [latex] x [/latex], and create a table of corresponding [latex] x [/latex] and [latex] y [/latex] values.

[latex] x [/latex]-values	[latex]2x-5[/latex]	[latex] y [/latex]-values	Ordered Pairs
[latex]0[/latex]	[latex]2(0)-5[/latex]	[latex]-5[/latex]	[latex] (0, -5) [/latex]
[latex]1[/latex]	[latex]2(1)-5[/latex]	[latex]-3[/latex]	[latex] (1, -3) [/latex]
[latex]2[/latex]	[latex]2(2)-5[/latex]	[latex]-1[/latex]	[latex] (2, -1) [/latex]
[latex]4[/latex]	[latex]2(4)-5[/latex]	[latex]3[/latex]	[latex] (4, 3) [/latex]
[latex]5[/latex]	[latex]2(5)-5[/latex]	[latex]5[/latex]	[latex] (5, 5) [/latex]

Plot the ordered pairs and draw a line through the points to obtain the graph.

Every point on the line is a solution to the equation [latex]y=2x–5[/latex]. You can substitute the values from the ordered pair [latex](\color{blue}{1},\color{Green}{−3})[/latex] into the equation and the equation will be satisfied because the point [latex] (1,-3) [/latex] is a point on the graph of [latex] y=2x-5 [/latex].

[latex]\begin{align}y&=2x-5\\ \color{Green}{-3}&\overset{\large{?}}{=}2\left(\color{blue}{1}\right)-5&& \color{blue}{\textsf{substitute}}\\-3&\overset{\large{?}}{=}2-5\\-3&=-3 && \color{blue}{\textsf{true statement}}\end{align}[/latex]

You can also substitute ANY of the other points on the line into the equation. Every point on the line is a solution to the equation [latex]y=2x–5[/latex].

To determine whether an ordered pair is a solution of a linear equation, substitute the [latex] x [/latex] and [latex] y [/latex] values into the equation. If this results in a true statement, then the ordered pair is a solution of the linear equation. If the result is not a true statement then the ordered pair is not a solution.

Below is a video further explaining how to determine if an ordered pair is a solution to a linear equation.

Intercepts

The intercepts are the points where the graph of the line intersects or crosses the horizontal and vertical axes. To help you remember what “intercept” means, think about the word “intersect.” The two words sound alike and in this case mean the same thing.

The straight line on the graph below intersects the two coordinate axes. The point where the line crosses the [latex] x [/latex]-axis is called the [latex] x [/latex]-intercept. The [latex] y [/latex]-intercept is the point where the line crosses the [latex] y [/latex]-axis. For this graph, the [latex] x [/latex]-intercept is the point [latex](−2,0)[/latex] and the [latex] y [/latex]-intercept is the point ([latex]0, 2[/latex]). The equation of the line below is [latex] x-y=-2 [/latex]. If you substitute [latex] 0 [/latex] in place of [latex] y [/latex], you find [latex] x [/latex] to be [latex] 2 [/latex], which is the [latex] x [/latex]-intercept of [latex] (0,2) [/latex]. If you substitute [latex] 0 [/latex] in place of [latex] y [/latex], you find [latex] x [/latex] to be [latex] -2 [/latex], which is the [latex] y [/latex]-intercept [latex] (-2,0) [/latex].

Notice that the [latex] y [/latex]-intercept always occurs where [latex]x=0[/latex], and the [latex] x [/latex]-intercept always occurs where [latex]y=0[/latex]. To find the [latex] x [/latex]– and [latex] y [/latex]-intercepts of a linear equation, you can substitute [latex]0[/latex] for [latex] y [/latex] and for [latex] x [/latex], respectively.

For example, the linear equation [latex]3y+2x=6[/latex] has an [latex] x [/latex]–intercept when [latex]\color{green}{y=0}[/latex]

[latex]\begin{align}3\left(\color{green}{0}\right)+2x&=6\\2x&=6\\x&=3\end{align}[/latex]

The [latex] x [/latex]-intercept is [latex](3,0)[/latex].

Likewise, the [latex] y [/latex]-intercept occurs when [latex]\color{blue}{x=0}[/latex]

[latex]\begin{align}3y+2\left(\color{blue}{0}\right)&=6\\3y&=6\\y&=2\end{align}[/latex]

The [latex] y [/latex]-intercept is [latex](0,2)[/latex].

Using Intercepts to Graph Lines

Intercepts can be used to graph linear equations. Once you have found the two intercepts, draw a line through them.

Do this with the equation [latex]3y+2x=6[/latex]. We found that the intercepts of the line this equation represents are [latex](0,2)[/latex] and [latex](3,0)[/latex]. Plot those two points and draw a line through them.

This video demonstrates graphing a linear equation using the intercepts.

Finding slope

Given two values for the input, [latex]{x}_{1}[/latex] and [latex]{x}_{2}[/latex], and two corresponding values for the output, [latex]{y}_{1}[/latex] and [latex]{y}_{2}[/latex], represented by a set of ordered pairs, [latex]\left({x}_{1}\text{, }{y}_{1}\right)[/latex] and [latex]\left({x}_{2}\text{, }{y}_{2}\right)[/latex], we can calculate the slope [latex]m[/latex] of a line through those points by

[latex]m=\dfrac{\text{change in output (rise)}}{\text{change in input (run)}}=\dfrac{\Delta y}{\Delta x}=\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex]

where [latex]\Delta y[/latex] is the vertical displacement and [latex]\Delta x[/latex] is the horizontal displacement.

The graph below indicates how the slope of the line between the points, [latex]\left({x}_{1,}{y}_{1}\right)[/latex] and [latex]\left({x}_{2,}{y}_{2}\right)[/latex] is calculated. The greater the absolute value of the slope, the “steeper” the line is.

The units for slope are always [latex]\dfrac{\text{units for the output}}{\text{units for the input}}[/latex]. Think of the units as the change in 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.

In the following video we show examples of how to find the slope of a line passing through two points.

In the next video, we show an example where we determine the increase in cost for producing solar panels given two data points.

The slope and y-intercept of a linear equation

Another way to graph a linear function is by using its slope [latex] m [/latex] and [latex] y [/latex]-intercept.

Consider the function [latex]f\left(x\right)=\dfrac{1}{2}x+1[/latex]. The function is in slope-intercept form, so the slope is [latex]m=\dfrac{1}{2}[/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The [latex] y [/latex]–intercept is the point on the graph where [latex]x=0[/latex]. The graph of this line crosses the [latex] y [/latex]-axis at [latex](0, 1)[/latex]. Begin graphing by plotting the point [latex](0, 1)[/latex]. The slope is [latex]m=\dfrac{\text{rise}}{\text{run}}[/latex], so for [latex]m=\dfrac{1}{2}[/latex], the “rise” is [latex]1[/latex] and the “run” from left to right is [latex]2[/latex]. Then starting from the [latex] y [/latex]–intercept [latex](0, 1)[/latex], “rise” [latex]1[/latex] unit and “run” [latex]2[/latex] units to the right. (This is equivalent to “run” [latex]2[/latex] units to the right and then “rise” [latex]1[/latex] unit.) Repeat until you have a few points and then draw a line through the points as shown in the graph below.

Example

Graph [latex]f\left(x\right)=-\dfrac{2}{3}x+5[/latex] using the [latex] y [/latex]–intercept and slope.

Show Solution

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

According to the equation for the function, the slope of the line is [latex]-\dfrac{2}{3}[/latex]. This tells us that for each vertical decrease in the “rise” of [latex]–2[/latex] units, the “run” increases by [latex]3[/latex] units in the horizontal direction. We can now graph the function by first plotting the y-intercept in the graph below. From the initial value [latex](0, 5)[/latex], we move down [latex]2[/latex] units and to the right [latex]3[/latex] units. (Which is equivalent to moving up [latex] 2 [/latex] units and left [latex] 3 [/latex] units.) We can extend the line to the left and right by using this relationship to plot additional points and then drawing a line through the points.

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

In the following video we show another example of how to graph a linear function given the y-intercepts and the slope.

In the following video, we show an example of how to write the equation of a line given its graph.

Point-Slope Form

If we have the slope of a line and a point on the line, we can write the equation of the line.

In this example, we will start with a given slope.

The following video shows an example of how to write the equation for a line given its slope and a point on the line.

Finding the Equation of a Line Given Two Points

If we start with two points, we can first find the slope of a line through those two points, and then find the equation of the line that passes through them.

Example

Write an equation for a linear function [latex] f [/latex] given its graph shown below.

Show Solution

Identify two points on the line such as [latex]\color{Green}{(0, 2)}[/latex] and [latex]\color{Blue}{(–2, –4)}[/latex]. Use the points to calculate the slope.

[latex]\begin{align}m&=\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\&=\dfrac{\color{Blue}{-4} - \color{Green}{2}}{\color{Blue}{-2} - \color{Green}{0}}\\&=\dfrac{-6}{-2}\\&=3\end{align}[/latex]

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

[latex]\begin{align}y-{y}_{1}&=m\left(x-{x}_{1}\right)\\y-\left(\color{Blue}{-4}\right)&=3\left(x-\left(\color{Blue}{-2}\right)\right)\\ y+4&=3\left(x+2\right)\end{align}[/latex]

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

[latex]\begin{align}y+4&=3\left(x+2\right)\\y+4&=3x+6\\ y&=3x+2\end{align}[/latex]

 

Examining the graph above, we see that the line crosses the [latex] y [/latex]-axis at the point [latex](0,2)[/latex] (the [latex] y [/latex]-intercept), so [latex]b=2[/latex] in the equation.

The next video shows another example of writing the equation of a line given two points on the line.

Vertical and Horizontal Lines

Vertical Lines

The equation of a vertical line is given as [latex] x=c [/latex] where [latex] c [/latex] is a constant. The slope of a vertical line is undefined, and regardless of the [latex] y [/latex]–value of any point on the line, the [latex] x [/latex]–coordinate of the point will be [latex] c [/latex].

Suppose that we want to find the equation of a line containing the following points: [latex]\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)[/latex], and [latex]\left(-3,5\right)[/latex]. First, we will find the slope, choosing two of the given points. In this case, we are choosing the points [latex] (-3,3) [/latex] and [latex] (-3,5) [/latex].

[latex]m=\dfrac{5 - 3}{-3-\left(-3\right)}=\dfrac{2}{0}[/latex]

Dividing by zero is undefined so the slope of the line through these points is undefined. Using point-slope form is not possible. Notice that all of the [latex] x [/latex]–coordinates in the given ordered pairs are the same. Then a vertical line through these points is [latex] x=-3 [/latex]. The line is on the graph below.

Horizontal Lines

The equation of a horizontal line is given as [latex] y=c [/latex] where [latex] c [/latex] is a constant. The slope of a horizontal line is zero, and for any [latex] x [/latex]–value of a point on the line, the corresponding [latex] y [/latex]–coordinate will be [latex] c [/latex].

Suppose we want to find the equation of a line that contains the following points: [latex]\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)[/latex], and [latex]\left(5,-2\right)[/latex]. We can use point-slope form to find the equation of the line. First, we find the slope using any two points on the line. In this case, we are choosing [latex] (-2,-2) [/latex] and [latex] (0,-2) [/latex].

[latex]\begin{align}m&=\dfrac{-2-\left(-2\right)}{0-\left(-2\right)}\\&=\dfrac{0}{2}\\&=0\end{align}[/latex]

Use any point for [latex]\left({x}_{1},{y}_{1}\right)[/latex] in point-slope form.

[latex]\begin{align}y-\left(-2\right)&=0\left(x - 3\right)\\y+2&=0\\ y&=-2\end{align}[/latex]

The graph (below) is a horizontal line [latex] y=-2 [/latex]. Notice that all of the [latex] y [/latex]–coordinates are the same in the list of ordered pairs.

The line [latex]x=−3[/latex] is a vertical line. The line [latex]y=−2[/latex] is a horizontal line.

Graphing a Linear Function: Another Approach

Another option for graphing linear functions is to use a vertical shift of the function.

Vertical Shift

In a linear function [latex]f\left(x\right)=mx+b[/latex], the [latex] b [/latex] acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in the figure below that adding a value of [latex] b [/latex] to the equation of [latex]f\left(x\right)=x[/latex] shifts the graph of [latex] f [/latex] a total of [latex] b [/latex] units up if [latex] b [/latex] is positive and [latex]|b|[/latex] units down if [latex] b [/latex] is negative. The graph below illustrates vertical shifts of the linear function [latex]f\left(x\right)=x[/latex].

In this next example, we will start with the function [latex] f(x)=\dfrac{1}{2}x [/latex] and use a vertical shift to graph [latex] f\left(x\right)=\dfrac{1}{2}x - 3 [/latex].

Example

Graph [latex]f\left(x\right)=\dfrac{1}{2}x - 3[/latex].

Show Solution

First graph [latex]y=\dfrac{1}{2}x[/latex] by starting at the [latex] y [/latex]-intercept of [latex] (0,0) [/latex] (this is also the [latex] x [/latex]-intercept). Then “rise” [latex] 1 [/latex] unit and “run” [latex] 2 [/latex] units to the right to plot a point at [latex] (2,1) [/latex]. Draw a line through those points.

Next, shift the line vertically down [latex] 3 [/latex] units to graph the line [latex] y=\dfrac{1}{2}x-3 [/latex].

Write the equation of a line given a point and a parallel or perpendicular line

The relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.

Let’s start with an example involving parallel lines. Parallel lines have the same slope.

Write the equation of a line given a point and a parallel line

Write the equation of a line given a point and a perpendicular line

When you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are perpendicular if the slope of one is the opposite reciprocal of the slope of the other. To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal.  In other words, flip it and change the sign.

Video: Write the equation of a line given a point and a perpendicular line (Alternate Method)

Write the equations of lines parallel and perpendicular to horizontal and vertical lines

Summary

The slope-intercept form of a linear equation is written as [latex]y=mx+b[/latex], where [latex] m [/latex] is the slope and [latex] y [/latex] is the value of [latex] y [/latex] at the [latex] y [/latex]-intercept, which can be written as [latex](0,b)[/latex]. When you know the slope and the [latex] y [/latex]-intercept of a line you can use the slope-intercept form to immediately write the equation of that line. The point-slope form, [latex]y-{y}_{1}=m\left(x-{x}_{1}\right)[/latex], can be used to write the equation of a line when you know the slope and a point on the line or when you know two points on the line.

When lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a 90° angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.