How do you write $x= 2y^2 - 28y + 116$ into vertex form?

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Answer 1 · 2015-05-31T18:51:41

I change the notation for easy understanding.

y = 2x^2 - 28x + 116

x of vertex: $x = (-b/2a) = 28/4 = 7$

y of vertex: $f(7) = 2.(49) - 28(7) + 116 = 98 - 196 + 116 = 18$

Vertex form $f(x) = 2(x - 7)^2 + 18$

Check by developing: f(x) = 2(x^2 - 14x + 49) + 18 =

= 2x^2 - 28x + 116. OK

How do you write x= 2y^2 - 28y + 116x= 2y^2 - 28y + 116 into vertex form?