How do you graph #y = x^2 - x - 2#?

1 Answer
Alan P.
Jul 23, 2015

Find #y# for some #x# values (including the vertex and the x and y intercepts, as a minimum); plot the corresponding points on the Cartesian plane; connect the points with a smooth parabolic curve.

Explanation:

The y-intercept (when #x=0#) occurs at #(0,-2)#

Since #(x^2-x-2)# can be factored as #(x-2)(x+1)#
the x-intercepts occur at #(2,0)# and #(-1,0)#

The vertex can be found by converting the given equation into vertex form:
#color(white)("XXXX")##y = (x-1/2)^2 -9/4#
or by taking the average x value for the x-intercepts [#(2+(-1))/2 = 1/2#] and evaluating the equation for #y#
The vertex occurs at #(1/2,-9/4)#

And the completed graph looks like:
graph{x^2-x-2 [-2.86, 4.933, -3.02, 0.874]}

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