Graphing slope-intercept form

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

Let's graph $y=2x+3$‍.

Recall that in the general slope-intercept equation $y=mx+b$‍, the slope is given by $m$‍ and the $y$‍-intercept is given by $b$‍. Therefore, the slope of $y=2x+3$‍ 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.

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 $(0+3,1+2)=(3,3)$‍.