What is the vertex form of #2y = 7x^2-5x+7#?

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Answer 1 · 2017-11-02T11:09:57

Vertex form of equation is #y =7/2(x-5/14)^2+3 3/56#

#2y =7x^2-5x+7 or y =7/2x^2-5/2x+7/2# or

#y =7/2(x^2-5/7x)+7/2# or

#y =7/2{x^2-5/7x+(5/14)^2}-7/2*(5/14)^2+7/2# or

#y =7/2{x^2-5/7x+(5/14)^2}-25/56+7/2#

#y =7/2(x-5/14)^2+171/56#. Comparing with vertex form of

equation #f(x) = a(x-h)^2+k ; (h,k)# being vertex we find

here #h=5/14 , k=171/56 or k = 3 3/56 #

So vertex is at #(5/14,3 3/56)# and vertex form of

equation is #y =7/2(x-5/14)^2+3 3/56# [Ans]