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)#