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 $y=2x+3$y, equals, 2, x, plus, 3.

Recall that in the general slope-intercept equation $y=\maroonC{m}x+\greenE{b}$y, equals, start color #ed5fa6, m, end color #ed5fa6, x, plus, start color #0d923f, b, end color #0d923f, the slope is given by $\maroonC{m}$start color #ed5fa6, m, end color #ed5fa6 and the $y$y-intercept is given by $\greenE{b}$start color #0d923f, b, end color #0d923f. Therefore, the slope of $y=\maroonC{2}x+\greenE{3}$y, equals, start color #ed5fa6, 2, end color #ed5fa6, x, plus, start color #0d923f, 3, end color #0d923f is $\maroonC{2}$start color #ed5fa6, 2, end color #ed5fa6 and the $y$y-intercept is $(0,\greenE{3})$left parenthesis, 0, comma, start color #0d923f, 3, end color #0d923f, right parenthesis.

In order to graph a line, we need two points on that line. We already know that $(0,\greenE{3})$left parenthesis, 0, comma, start color #0d923f, 3, end color #0d923f, right parenthesis is on the line.

Additionally, because the slope of the line is $\maroonC{2}$start color #ed5fa6, 2, end color #ed5fa6, we know that the point $(0\maroonC{+1},\greenE{3}\maroonC{+2})=(1,5)$left parenthesis, 0, start color #ed5fa6, plus, 1, end color #ed5fa6, comma, start color #0d923f, 3, end color #0d923f, start color #ed5fa6, plus, 2, end color #ed5fa6, right parenthesis, equals, left parenthesis, 1, comma, 5, right parenthesis is also on the line.

Let's graph $y=\maroonC{\dfrac{2}{3}}x\greenE{+1}$y, equals, start color #ed5fa6, start fraction, 2, divided by, 3, end fraction, end color #ed5fa6, x, start color #0d923f, plus, 1, end color #0d923f.

As before, we can tell that the line passes through the $y$y-intercept $(0,\greenE{1})$left parenthesis, 0, comma, start color #0d923f, 1, end color #0d923f, right parenthesis, and through an additional point $\left(0\maroonC{+1},\greenE{1}\maroonC{+\dfrac{2}{3}}\right)=\left(1,1\dfrac{2}{3}\right)$left parenthesis, 0, start color #ed5fa6, plus, 1, end color #ed5fa6, comma, start color #0d923f, 1, end color #0d923f, start color #ed5fa6, plus, start fraction, 2, divided by, 3, end fraction, end color #ed5fa6, right parenthesis, equals, left parenthesis, 1, comma, 1, start fraction, 2, divided by, 3, end fraction, right parenthesis.

While it is true that the point $\left(1,1\dfrac{2}{3}\right)$left parenthesis, 1, comma, 1, start fraction, 2, divided by, 3, end fraction, right parenthesis 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 $\maroonC{\dfrac{2}{3}}$start color #ed5fa6, start fraction, 2, divided by, 3, end fraction, end color #ed5fa6, increasing $x$x by $\maroonC{3}$start color #ed5fa6, 3, end color #ed5fa6 units will cause $y$y to increase by $\maroonC{2}$start color #ed5fa6, 2, end color #ed5fa6 units.

This gives us the additional point $(0\maroonC{+3},\greenE{1}\maroonC{+2})=(3,3)$left parenthesis, 0, start color #ed5fa6, plus, 3, end color #ed5fa6, comma, start color #0d923f, 1, end color #0d923f, start color #ed5fa6, plus, 2, end color #ed5fa6, right parenthesis, equals, left parenthesis, 3, comma, 3, right parenthesis.