How do you graph using slope and intercept of #- 5x + y =0#?

1 Answer
Dec 21, 2015

You have to move some things around in the equation to put it into slope-intercept form, then you can graph it.

Explanation:

#-5x+y=0#

Slope intercept form is: #y=mx+b#

First, you need to get the #y# alone on the left side of the equation.
#-5x+y=0#
#-5x+y+5x=0+5x#
#y=0+5x#

Now, the equation is almost correct and ready to be graphed, but for it to be technically correct, the place of the #0# and the #5x# need to be switched.
#y=5x+0#

So, you know that the y-intercept is #0#, which means that it is at the point #(0,0)#
Plot a point at #(0,0)#

Then, since you have a slope of #5#, that means you have a rise, or change in the #y# value, of #+5#, and a run, or change in the #x# value, of #+1#

Starting at point #(0,0)#, plot at point five up, and one to the right. So, that point would be at #(1,5)#. Then, from #(1,5)#, go up five, and to the right one, so you'll be at point #(2,10)#. Then, draw a line through your points.