Graphing Linear Functions

We previously 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.

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 [latex]y[/latex]-intercept and slope. The third is applying transformations to the identity function [latex]f\left(x\right)=x[/latex].

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 of the function.

For example, given the function [latex]f\left(x\right)=2x[/latex], we might use the input values [latex]1[/latex] and [latex]2[/latex]. Evaluating the function for an input value of [latex]1[/latex] yields an output value of [latex]2[/latex] which is represented by the point [latex](1, 2)[/latex]. Evaluating the function for an input value of [latex]2[/latex] yields an output value of [latex]4[/latex] which is represented by the point [latex](2, 4)[/latex]. Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

Graph [latex]f\left(x\right)=-\frac{2}{3}x+5[/latex] by plotting points.

Show Solution

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

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

[latex]\begin{array}{llllll}x=0& & f\left(0\right)=-\frac{2}{3}\left(0\right)+5=5\Rightarrow \left(0,5\right)\\ x=3& & f\left(3\right)=-\frac{2}{3}\left(3\right)+5=3\Rightarrow \left(3,3\right)\\ x=6& & f\left(6\right)=-\frac{2}{3}\left(6\right)+5=1\Rightarrow \left(6,1\right)\end{array}[/latex]

Plot the coordinate pairs and draw a line through the points. The graph below is of the function [latex]f\left(x\right)=-\frac{2}{3}x+5[/latex].

 

Analysis of the Solution

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.

Graphing a Linear 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 [latex]y[/latex]-intercept which is the point at which the input value is zero. To find the [latex]y[/latex]-intercept, we can set [latex]x=0[/latex] in the equation.

The other characteristic of the linear function is its slope [latex]m[/latex], which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a linear 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, between any two points by the horizontal difference, or run. The slope of a linear function will be the same between any two points. We encountered both the [latex]y[/latex]-intercept and the slope in linear functions.

Let’s consider the following function.

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

The slope is [latex]\frac{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 when [latex]x= 0[/latex]. The graph crosses the [latex]y[/latex]-axis at [latex](0, 1)[/latex]. Now we know the slope and the [latex]y[/latex]-intercept. We can begin graphing by plotting the point [latex](0, 1)[/latex] We know that the slope is rise over run, [latex]m=\frac{\text{rise}}{\text{run}}[/latex]. From our example, we have [latex]m=\frac{1}{2}[/latex], which means that the rise is [latex]1[/latex] and the run is [latex]2[/latex]. Starting from our [latex]y[/latex]-intercept [latex](0, 1)[/latex], we can rise [latex]1[/latex] and then run [latex]2[/latex] or run [latex]2[/latex] and then rise [latex]1[/latex]. We repeat until we have multiple points, and then we draw a line through the points as shown below.

Graph [latex]f\left(x\right)=-\frac{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]-\frac{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 [latex]y[/latex]-intercept. From the initial value [latex](0, 5)[/latex] we move down [latex]2[/latex] units and to the right [latex]3[/latex] units. We can extend the line to the left and right by repeating, and then draw a line through the points.

 

Analysis of the Solution

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