What is the vertex form of #y= 7x^2+2x+12 #?

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Answer 1 · 2017-05-16T05:25:00

#y=7(x+1/7)^2+83/7#

Vertex form of equation is #y=a(x-h)^2+k# and #(h,k)# is the vertex.

We have #y=7x^2+2x+12#

= #7(x^2+2xx1/7xx x+(1/7)^2-(1/7)^2)+12#

= #7(x+1/7)^2-7(1/7)^2+12#

= #7(x+1/7)^2-1/7+12#

= #7(x+1/7)^2+83/7#

i.e. #y=7(x+1/7)^2+83/7#