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### Course: Algebra (all content) > Unit 3

Lesson 6: Graphing slope-intercept equations# Graphing slope-intercept form

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 $y={m}x+{b}$ , the slope is given by ${m}$ and the $y$ -intercept is given by ${b}$ .
Therefore, the slope of $y={2}x+{3}$ is ${2}$ and the $y$ -intercept is $(0,{3})$ .

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

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

## Check your understanding

## Graphing lines with fractional slope

Let's graph $y={{\displaystyle \frac{2}{3}}}x{+1}$ .

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

While it is true that the point $(1,1{\displaystyle \frac{2}{3}})$ 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 ${{\displaystyle \frac{2}{3}}}$ , increasing $x$ by ${3}$ units will cause $y$ to increase by ${2}$ units.

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

## Check your understanding

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