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+3y=2x+3.
Recall that in the general slope-intercept equation y=mx+by=\maroonC{m}x+\greenE{b}, the slope is given by m\maroonC{m} and the yy-intercept is given by b\greenE{b}. Therefore, the slope of y=2x+3y=\maroonC{2}x+\greenE{3} is 2\maroonC{2} and the yy-intercept is (0,3)(0,\greenE{3}).
In order to graph a line, we need two points on that line. We already know that (0,3)(0,\greenE{3}) is on the line.
Additionally, because the slope of the line is 2\maroonC{2}, we know that the point (0+1,3+2)=(1,5)(0\maroonC{+1},\greenE{3}\maroonC{+2})=(1,5) is also on the line.
The fact that the slope of the line is 2\maroonC{2} tells us that when xx increases by 1\maroonC{1} unit, yy increases by 2\maroonC{2} units.
Increasing xx by 1\maroonC{1} unit and yy by 2\maroonC{2} units from (0,3)(0,3) gives us (0+1,3+2)(0\maroonC{+1},3\maroonC{+2}), which is (1,5)(1,5).

Check your understanding

Problem 1
Graph y=3x1y=3x-1.
Problem 2
Graph y=4x+5y=-4x+5.

Graphing lines with fractional slope

Let's graph y=23x+1y=\maroonC{\dfrac{2}{3}}x\greenE{+1}.
As before, we can tell that the line passes through the yy-intercept (0,1)(0,\greenE{1}), and through an additional point (0+1,1+23)=(1,123)\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 (1,123)\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 23\maroonC{\dfrac{2}{3}}, increasing xx by 3\maroonC{3} units will cause yy to increase by 2\maroonC{2} units.
Remember that slope is the ratio of the change in yy over the change in xx.
Slope=Change in yChange in x23=Change in yChange in xSlope=2323=Change in y3Change in x=32=Change in yMultiply by 3\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 xx is the same as the denominator of the slope, then the change in yy must be the same as the numerator of the slope.
This can be generalized as follows: For a slope of ab\dfrac{a}{b}, an increase in xx by bb units causes an increase in yy by aa units.
This gives us the additional point (0+3,1+2)=(3,3)(0\maroonC{+3},\greenE{1}\maroonC{+2})=(3,3).

Check your understanding

Problem 3
Graph y=34x+2y=\dfrac{3}{4}x+2.
Problem 4
Graph y=32x+3y=-\dfrac{3}{2}x+3.