How do you write $f(x)= 6x-x^2$ in vertex form?

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Answer 1 · 2018-04-09T04:53:45

The general vertex form is $y = a(x-h)^2+ k$ where $(h,k)$ is the vertex.

$f(x) = 6x - x^2$
$f(x) = -x^2 + 6x - 9 + 9$
$f(x) = -(x^2 - 6x + 9) - 9$

Vertex form:
$f(x) = -(x-3)^2 - 9$

Vertex:
$(3,-9)$

How do you write f(x)= 6x-x^2f(x)= 6x-x^2 in vertex form?