Graph a Linear Function - Pre-Calculus

Which of the following could be the function modeled by this graph?

We can begin here by trying to identify a couple points on the graph

We can see that it crosses the y-axis at

Therefore, not only do we have a point, we have the y-intercept. This tells us that the equation of the line needs to have a in it somewhere. Eliminate any option that do not have this feature.

Next, find the slope by counting up and over from the y-intercept to the next clear point.

It seems like the line goes up 5 and right 1 to the point

This means we have a slope of 5, which means our equation must look like this:

For the linear function in point-slope form

The slope is equal to

For this problem

we get

By definition, the y-intercept is the point on the line that crosses the y-axis. This can be found by substituting into the equation. When we do this with our equation,

.

Alternatively, you can remember form, a general form for a line in which is the slope and is the y-intercept.

Recall slope-intercept form, or . In this form, is the slope and is the y-intercept. Given our equation above, the slope must be the coefficient of the x, which is .

The x-intercept of an equation is the point at which the line crosses the x-axis. Thus, we can find the x-intercept by plugging in . When we do this with our equation:

Thus, our x-intercept is the point .