How do you graph #y=2x^2-8x-1# by plotting points?

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Answer 1 · 2015-12-05T16:17:04

Make a table of #x# and #y#.

Try plugging in an #x#-value: a good one to start with is always #0#:

#y=2(0)^2-8(0)-1#
#y=0-0-1#
#y=-1#

Thus, when #x=0#, #y=-1#, so we can plot the point #(0,-1)# on our graph.

Continue to find points and graph them:
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Eventually, you should see the shape of the graph starting to form.

Connect the dots, and you should see this:

graph{2x^2-8x-1 [-26.94, 24.38, -11.3, 14.36]}