How do you graph #y=-2+8x#?

1 Answer
Apr 25, 2017

#x# intercept is #(1/4,0)#

#y# intercept is #(0,-2)#

slope is #8#

Explanation:

We need a few pieces of information to be able to graph a function. We need the #x# and #y# intercepts and the slope.

We can use slope-intercept form to help us find two of these components. Slope-line form is #y=mx+b#, where #m# is slope and #b# is the #y# intercept.

In our case, #m# is #8#, so our slope is #8#, and our #y# intercept is #-2#.

So, we have our slope and #y# intercept, but we still need our #x# intercept. To find that, we set #y# to #0# and solve for #x#:

#0=-2+8x#

add #2# to both sides

#2=8x#

divide by #8#

#1/4=x#

So, our #x# intercept is #(1/4,0)#, our #y# intercept is #(0,-2)#, and our slope is #8#, or up #8#, over #1#

The final graph should look like this:
graph{y=-2+8x}