How do you find the vertex of a quadratic equation #y=(x+8)^2-2#?

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Answer 1 · 2015-11-28T20:15:24

#x_("vertex")=-8#

I will let you find #y_("vetex")# by sustitution

Expanding the brackets gives:
#y=x^2+16x+64-2#
#y=x^2+16x+64#

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The equation given in the question is a quadratic that is already in vertex form (completing the square).

Consider the +8 from #(x+8)^2#

The #x_("vertex") " should be at "(-1)xx8=-8#
Lets have a look!
Tony B
Looks good!

Now all you have to do is substitute #x=(-8)# in the equation to find the value of y.

I will let you do that!