Graph Linear Functions Using Slope and y-Intercept

Learning Outcome

Graph a linear function using the slope and y-intercept

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

Let us consider the following function.

$f\left(x\right)=\dfrac{1}{2}x+1$

The function is in slope-intercept form, so the slope is $\dfrac{1}{2}$ . 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 $x=0$ . The graph crosses the y-axis at $(0, 1)$ . Now we know the slope and the y-intercept. We can begin graphing by plotting the point $(0, 1)$ . We know that the slope is rise over run, $m=\dfrac{\text{rise}}{\text{run}}$ . From our example, we have $m=\dfrac{1}{2}$ , which means that the rise is $1$ and the run is $2$ . So starting from our y-intercept $(0, 1)$ , 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 the graph below.

graph of the line y = (1/2)x +1 showing the "rise", or change in the y direction as 1 and the "run", or change in x direction as 2, and the y-intercept at (0,1)

A General Note: Graphical Interpretation of a Linear Function

In the equation $f\left(x\right)=mx+b$

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

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

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

Evaluate the function at an input value of zero to find the y-intercept.
Identify the slope.
Plot the point represented by the y-intercept.
Use $\dfrac{\text{rise}}{\text{run}}$ to determine at least two more points on the line.
Sketch the line that passes through the points.

Example

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

Show Solution

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

According to the equation for the function, the slope of the line is $-\dfrac{2}{3}$ . This tells us that for each vertical decrease in the “rise” of $–2$ units, the “run” increases by $3$ 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 $(0, 5)$ , we move down $2$ units and to the right $3$ 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.

graph of the line y = (-2/3)x + 5 showing the change of -2 in y and change of 3 in x.

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

Try It

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

In the last example, we will show how to graph another linear function using the slope and y-intercept.

Example

Graph $f\left(x\right)=-\dfrac{3}{4}x+6$ using the slope and y-intercept.

Show Solution

The slope of this function is $-\dfrac{3}{4}$ and the y-intercept is $(0,6)$ We can start graphing by plotting the y-intercept and counting down three units and right $4$ units. The first stop would be $(4,3)$ , and the next stop would be $(8,0)$ .

Screen Shot 2016-07-13 at 1.35.32 PM

Module 5: Linear Functions