# How do you find the critical points for  y = x^3 + 12x^2 + 6x + 8 ?

Since ${\lim}_{x \to \pm \infty} f \left(x\right) = \pm \infty$ there is no global maximum or minimum.
It has a local maximum at $x = - 4 - \sqrt{14}$ which is
$f \left(- 4 - \sqrt{14}\right) = 28 \left(4 + \sqrt{14}\right)$
and it has local minimum at $x = - 4 + \sqrt{14}$ which is
$f \left(- 4 + \sqrt{14}\right) = - 28 \left(\sqrt{14} - 4\right)$