# Graphing slope-intercept form

CCSS Math: 8.F.A.1, 8.F.A.3, HSF.IF.C.7, HSF.IF.C.7a

Learn how to graph lines whose equations are given in the slope-intercept form y=mx+b.

If you haven't read it yet, you might want to start with our introduction to slope-intercept form.

## Graphing lines with integer slopes

Let's graph $y=2x+3$.

Recall that in the general slope-intercept equation $y=\maroonC{m}x+\greenE{b}$, the slope is given by $\maroonC{m}$ and the $y$-intercept is given by $\greenE{b}$.
Therefore, the slope of $y=\maroonC{2}x+\greenE{3}$ is $\maroonC{2}$ and the $y$-intercept is $(0,\greenE{3})$.

In order to graph a line, we need two points on that line. We already know that $(0,\greenE{3})$ is on the line.

Additionally, because the slope of the line is $\maroonC{2}$, we know that the point $(0\maroonC{+1},\greenE{3}\maroonC{+2})=(1,5)$ is also on the line.

## Check your understanding

## Graphing lines with fractional slope

Let's graph $y=\maroonC{\dfrac{2}{3}}x\greenE{+1}$.

As before, we can tell that the line passes through the $y$-intercept $(0,\greenE{1})$, and through an additional point $\left(0\maroonC{+1},\greenE{1}\maroonC{+\dfrac{2}{3}}\right)=\left(1,1\dfrac{2}{3}\right)$.

While it is true that the point $\left(1,1\dfrac{2}{3}\right)$ is on the line, we can't plot points with fractional coordinates as precisely as we draw points with integer coordinates.

We need a way to find another point on the line whose coordinates are integers. To do that, we use the fact that in a slope of $\maroonC{\dfrac{2}{3}}$, increasing $x$ by $\maroonC{3}$ units will cause $y$ to increase by $\maroonC{2}$ units.

This gives us the additional point $(0\maroonC{+3},\greenE{1}\maroonC{+2})=(3,3)$.