# How do you graph a line using slope-intercept form?

Jan 30, 2015

The equation of a line in explicit form is:

$y = m x + q$, where $m$ is the slope and $q$ the y-intercept.

It is easier to show the procedure with some example:

$y = 2$: this line is parallel to the x-axis and it passes from the point $P \left(0 , 2\right)$.

$x = 3$: this line is parallel to the y-axis and it passes from the point $P \left(2 , 0\right)$.

$y = x + 1$: this line is parallel to the bisector of the I and III quadrants and it passes from the point $P \left(0 , 1\right)$.

graph{x+1 [-10, 10, -5, 5]}

$y = - x - 1$: this line is parallel to the bisector of the II and IV quadrants and it passes from the point $P \left(0 , - 1\right)$.

graph{-x-1 [-10, 10, -5, 5]}

$y = \frac{2}{3} x + 1$: we have to find the point $P \left(0 , 1\right)$, from this point we have to "count" 3 units to the right and then 2 units to the up, so we can find the point Q(3,3), then we have to join the two point found.

graph{2/3x+1 [-10, 10, -5, 5]}

$y = - \frac{1}{2} x - 1$: we have to find the point $P \left(0 , - 1\right)$, from this point we have to "count" 2 units to the left and then 2 units to the up, so we can find the point Q(-2,0), then we have to join the two point found.

graph{-1/2x-1 [-10, 10, -5, 5]}

The difference in these two last examples is the "choice" of the "right" and the "left". Right, if the $m$ is positive; left, if the $m$ is negative.