How do you write #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)#

he parabola equation is of the form

#y=a(x-h)^2+k#

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

Rearrange your equation

#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]}