What is the vertex of # y= x^2-x-16+(x-1)^2#?

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Answer 1 · 2017-12-13T20:23:39

First, expand the expression and combine like terms:

#x^2-x-16+(x-1)^2#

#\implies x^2-x-16+(x^2-2x+1)#

#\implies x^2+x^2-x-2x-16+1#

#\implies 2x^2-3x-15#

Now that's in the form #ax^2+bx+c#, the vertex's #x#-coordinate is #\frac{-b}{2a}#.

#\implies \frac{3}{4}#

Plug that into the original equation to find the #y#-coordinate:

#2x^2-3x-15#

#\implies 2(3/4)^2-3(3/4)-15#

#\implies 9/8-9/4-15/1#

#implies -16.125#

I'm in class rn and will finish this later. Sorry. :/