What is the vertex of $y= -x^2 + 12x + 36$ ?

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Answer 1 · 2016-03-28T06:51:42

$(6,72)$

$y = -(x-6)^2 + 72$
graph{-(x-6)^2 + 72 [-20, 20, -80, 80]}
The vertex is the topmost point of the parabola.

Maximum occurs when $x=6$ , and $y=72$ .

The vertex is $(6,72)$ .

What is the vertex of y= -x^2 + 12x + 36 y= -x^2 + 12x + 36?