How do you write $f (x) = - x^{2} + 4 x + 6$ $f(x)=-x^2+4x+6$ in vertes form?

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Answer 1 · 2016-05-30T09:49:43

The vertex is at $(2, 10)$ $(2,10)$

he parabola equation is of the form

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

The vertex is at point $(h, k)$ $(h,k)$

Rearrange your equation

$y = - x^{2} + 4 x + 6$ $y=-x^2+4x+6$

$y=underbrace(-x^2+4x-4)+10$

$y= -(x^2-4x+4)+10$

$y=-(x-2)^2+10$

The vertex is at $(2,10)$

graph{-x^2+4x+6 [-10.92, 9.08, -12.24, -2.24]}

How do you write f(x)=-x^2+4x+6f(x)=−x2+4x+6f(x)=-x^2+4x+6 in vertes form?