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

May 30, 2016

The vertex is at $\left(2 , 10\right)$

Explanation:

he parabola equation is of the form

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

The vertex is at point $\left(h , k\right)$

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

$y = \underbrace{- {x}^{2} + 4 x - 4} + 10$

$y = - \left({x}^{2} - 4 x + 4\right) + 10$

$y = - {\left(x - 2\right)}^{2} + 10$

The vertex is at $\left(2 , 10\right)$

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