# Slope-Intercept Form

If you know the
slope
$m$
, and
$y$
-intercept
$(0,b)$
of a line (the point where the line crosses the
$y$
-axis), you can write the equation of the line in
**
slope-intercept form
**
.

$y=mx+b$

(You can think of this as a special case of the point-slope form of the equation where $({x}_{1},{y}_{1})$ is the point $(0,b)$ .)

**
Example 1
**
:

Find an equation of the line in slope-intercept form with slope $3$ and $y$ -intercept $(0,-2)$ .

$y=3x-2$ .

**
Example 2
**
:

Find an equation of the line in slope-intercept form with $y$ -intercept $(0,4)$ and passing through the point $(2,9)$ .

First, find the slope of the line: $m=\frac{9\text{\hspace{0.17em}}-\text{\hspace{0.17em}}4}{2\text{\hspace{0.17em}}-\text{\hspace{0.17em}}0}=\frac{5}{2}$

Then, write the equation: $y=\frac{5}{2}x+4$

Equations in this form are easy to graph, since the slope of the line is
**
$m$
**
and the
$y$
-intercept of the line is
**
$b$
**
.

**
Example 3
**
:

Rewrite the equation $y-1=-3(x+2)$ in slope-intercept form.

The equation is already in point-slope form; we know that $-3$ is the slope.

Expand the right side using the distributive property.

$y-1=-3x-6$

Add $1$ to both side.

$y=-3x-5$

Now we have the equation in slope-intercept form.

**
Example 4
**
:

Graph $y=-2x+3$ .

Since the equation is given in slope-intercept form, we know immediately that the line crosses the $y$ -axis at $(0,3)$ and has slope $-2$ . We can quickly use the slope to find a second point $(1,1)$ , and graph the line.