What is the vertex of #y=-x^2+4x+12#? (This replaces the same question that I accidentally deleted before my first cup of coffee).

1 Answer
Jun 29, 2015

The vertex of #-x^2+4x+12# is at #(2,16)#

Explanation:

By rewriting #y=-x^2+4x+12#
into "vertex form": #y=m(x-a)^2+b# (with vertex at #(a,b)#)
we can simply "read off" the vertex values.

#y=-x^2+4x+12#
#color(white)("XXXX")#extract #m#
#y=(-1)(x^2-4x-12)#
#color(white)("XXXX")#complete the square
#y=(-1)(color(blue)(x^2-4x+4) -12 -4)#
#color(white)("XXXX")#rewrite as a square plus an external term
#y=(-1)(x-2)^2 +16#

This is in vertex form with the vertex at #(2,16)#