# How do you find the max and min of y=-x^2-8x+2 by completing the square?

May 31, 2015

$- {x}^{2} - 8 x + 2 = - \left({x}^{2} + 8 x - 2\right)$

$= - \left({\left(x + 4\right)}^{2} - 16 - 2\right)$

$= - \left({\left(x + 4\right)}^{2} - 18\right)$

$= 18 - {\left(x + 4\right)}^{2}$

Being the square of a real number ${\left(x + 4\right)}^{2} \ge 0$

So the maximum value $18$ occurs when ${\left(x + 4\right)}^{2} = 0$ (that is when $x = - 4$)

The value is unbounded in the other direction.

$y = - {x}^{2} - 8 x + 2 \to - \infty$ as $x \to \infty$ or $x \to - \infty$