How do you find the max and min of #y=-x^2+6# by completing the square?

1 Answer
May 9, 2015

"Completing the squares" is a bit unusual for the given equation:
#y=-x^2+6#
since in "completed square form" it would be:
#y = (-1)(x^2-0x+0)+6#
or
#y=(-1)(x-0)^2+6#

This is also the vertex form;
so the vertex is at #(0,6)#

Because of the #(-1)# factor we know that the parabola opens downward #rarr# the vertex is at a maximum.
As the magnitude of #x# increases the value of #y# decreases without bound.

So the maximum value is #6# and the minimum is #-oo#