# How do you find the local max and min for f(x)= x^2 + 8x -12?

Jun 22, 2016

${y}_{\min} = - 28$ at $x = - 4$

#### Explanation:

calculus is OTT for this

So

$y = {x}^{2} + 8 x - 12 = {\left(x + 4\right)}^{2} - 16 - 12$
$= {\left(x + 4\right)}^{2} - 28$

${\left(x + 4\right)}^{2} \setminus \ge 0$ so ${y}_{\min} = - 28$ at $x = - 4$