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

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:
$y = \left(- 1\right) \left({x}^{2} - 0 x + 0\right) + 6$
or
$y = \left(- 1\right) {\left(x - 0\right)}^{2} + 6$

This is also the vertex form;
so the vertex is at $\left(0 , 6\right)$

Because of the $\left(- 1\right)$ factor we know that the parabola opens downward $\rightarrow$ 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 $- \infty$