How do you graph #f(x)=1/4x^2-2x-12# and identify the x intercepts, vertex?

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Answer 1 · 2018-02-02T08:05:43

Refer Explanation section.

Given -

#y=1/2 x^2-2x-12#

Vertex -
x-coordinate

#(-b)/(2a)=(-(-2))/(2 xx0.5)=2/1=2#

Y coordinate
At #x=2#

#y=1/2(2)^2-2(2)-12=4/2-4-12=2-4-12=-14#

Vertex #(2,-14)#

Take a few points on either side of #x=2#. Find the corresponding y values. Tabulate them.
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Plot the points and join with a smooth curve.

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Its two x-intercepts enter image source here are given in a separate graph