# How do you find the critical points for y = 2x^3 + 3x^2 - 12x + 4?

May 23, 2015

The critical points are $x = - 2$ and $x = 1$.

You evaluate the first derivative and find its zeroes:

$y ' = 2 \cdot 3 {x}^{2} + 3 \cdot 2 x - 12 \cdot 1 + 0 = 6 {x}^{2} + 6 x - 12$
$y ' = 6 {x}^{2} + 6 x - 12$
And now we can use the quadratic formula
$\Delta = {6}^{2} - 4 \cdot 6 \cdot \left(- 12\right) = 324$
$\sqrt{\Delta} = 18$
${x}_{1} = \frac{- 6 - 18}{2 \cdot 6} = \frac{- 24}{12} = - 2$
${x}_{2} = \frac{- 6 + 18}{2 \cdot 6} = \frac{12}{12} = 1$