# Graphing slope-intercept form

CCSS Math: 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.
The fact that the slope of the line is $\maroonC{2}$ tells us that when $x$ increases by $\maroonC{1}$ unit, $y$ increases by $\maroonC{2}$ units.
Increasing $x$ by $\maroonC{1}$ unit and $y$ by $\maroonC{2}$ units from $(0,3)$ gives us $(0\maroonC{+1},3\maroonC{+2})$, which is $(1,5)$.

Problem 1
Graph $y=3x-1$.
Problem 2
Graph $y=-4x+5$.

## 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.
Remember that slope is the ratio of the change in $y$ over the change in $x$.
\begin{aligned}\text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x}\\\\\\\\ \dfrac{2}{3}&=\dfrac{\text{Change in }y}{\text{Change in }x}&\small\gray{\text{Slope}=\dfrac{2}{3}}\\\\\\\\\\ \dfrac{2}{3}&=\dfrac{\text{Change in }y}{3}&\small\gray{\text{Change in }x=3}\\\\\\ 2&=\text{Change in }y&\small\gray{\text{Multiply by 3}} \end{aligned}
In other words, if the change in $x$ is the same as the denominator of the slope, then the change in $y$ must be the same as the numerator of the slope.
This can be generalized as follows: For a slope of $\dfrac{a}{b}$, an increase in $x$ by $b$ units causes an increase in $y$ by $a$ units.
This gives us the additional point $(0\maroonC{+3},\greenE{1}\maroonC{+2})=(3,3)$.

Graph $y=\dfrac{3}{4}x+2$.
Graph $y=-\dfrac{3}{2}x+3$.