What is the vertex form of y= x^2+x-12y=x2+x−12?

1 Answer
George C.
Dec 19, 2015

Complete the square to find:

y = 1(x-(-1/2))^2+(-49/4)y=1(x−(−12))2+(−494)

in vertex form

Explanation:

Complete the square as follows:

y = x^2+x-12y=x2+x−12

= x^2+x+1/4-1/4-12=x2+x+14−14−12

= (x+1/2)^2-49/12=(x+12)2−4912

That is:

y = 1(x-(-1/2))^2+(-49/4)y=1(x−(−12))2+(−494)

This is in vertex form:

y = a(x-h)^2+ky=a(x−h)2+k

with a=1a=1, h=-1/2h=−12 and k=-49/4k=−494

so the vertex is at (h, k) = (-1/2, -49/4)(h,k)=(−12,−494)

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