Slope-Intercept Form

Key Questions

How do you find the "m" and "b" of any linear equation?
Answer:

$m$ $m$ is the slope, while $b$ $b$ is the y-intercept.
Explanation:
Any linear equation has the form of

$y = m x + b$ $y=mx+b$

$m$ $m$ is the slope of the equation

$b$ $b$ is the y-intercept

The slope of the line, $m$ $m$ , is found by

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m=(y_2-y_1)/(x_2-x_1)$

where $(x_{1}, y_{1})$ $(x_1,y_1)$ and $(x_{2}, y_{2})$ $(x_2,y_2)$ are the coordinates of any two points in the line.

The y-intercept, $b$ $b$ , is found by plugging in $x = 0$ $x=0$ into the equation, which results in $y = b$ $y=b$ , and therefore is the y-intercept.

In some cases, if the equation is already arranged for you nicely, like $y = 3 x + 5$ $y=3x+5$ , we can easily find the y-intercept for this line, which is $5$ $5$ .

Other times, the equation might not be arranged nicely, with cases such as $\frac{1}{2} x + 3 y = 5$ $1/2x+3y=5$ , in which we solve for the y-intercept:

$\frac{1}{2} x + 3 y = 4$ $1/2x+3y=4$

$3 y = 4 - \frac{1}{2} x$ $3y=4-1/2x$

$y = \frac{- \frac{1}{2} x + 4}{3}$ $y=(-1/2x+4)/3$

$y = - \frac{1}{6} x + \frac{4}{3}$ $y=-1/6x+4/3$

So, the y-intercept of this line is $\frac{4}{3}$ $4/3$ .
Nam D. · 1 · Mar 19 2018
How do you determine the slope and y intercept when given a graph?

The $y$ $y$ -intercept $b$ $b$ can be found by reading the $y$ $y$ -axis where the graph hits the y-axis, and the slope $m$ $m$ can be found by finding any two distinct points $(x_{1}, y_{1})$ $(x_1,y_1)$ and $(x_{2}, y_{2})$ $(x_2,y_2)$ on the graph, and using the slope formula below.

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m={y_2-y_1}/{x_2-x_1}$ .

I hope that this was helpful.

Wataru · 2 · Oct 29 2014
What is Slope-Intercept Form?

Answer:

$y = m x + b$ $y = mx + b$

Where:
$m$ $m$ is the slope of the line.
$b$ $b$ is the y-intercept of the line.

Explanation:

Consider $y = x$ $y = x$

graph{y=x [-10, 10, -5, 5]}

In this equation, the coefficient to $x$ $x$ is 1 and our y-intercept is 0.

We could think of that equation as looking like:

$y = 1 x + 0$ $y = 1x + 0$

Notice that the graphed line has a "rise-over-run" of $\frac{1}{1}$ $1/1$ which is just 1 and the line passing through the y-axis at $y = 0$ $y=0$

Matt B. · 1 · Apr 5 2018