What is the vertex form of $y= x^2+x-12$ ?

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Answer 1 · 2015-12-19T00:23:51

Complete the square to find:

$y = 1(x-(-1/2))^2+(-49/4)$

in vertex form

Complete the square as follows:

$y = x^2+x-12$

$= x^2+x+1/4-1/4-12$

$= (x+1/2)^2-49/12$

That is:

$y = 1(x-(-1/2))^2+(-49/4)$

This is in vertex form:

$y = a(x-h)^2+k$

with $a=1$ , $h=-1/2$ and $k=-49/4$

so the vertex is at $(h, k) = (-1/2, -49/4)$

What is the vertex form of y= x^2+x-12y= x^2+x-12?