# How do you use the first and second derivatives to sketch y = x^3 - 12x - 12?

Feb 1, 2016

#### Explanation:

$y = {x}^{3} - 12 x - 12$
First derivative: $y ' = 3 {x}^{2} - 12$
Second derivative: $y ' ' = 6 x$

Using $y ' = 3 {x}^{2} - 12$
$3 \left({x}^{2} - 4\right)$
$3 \left(x - 2\right) \left(x + 2\right)$
$x = - 2 , 2$
Sub x values into $y = {x}^{3} - 12 x - 12$, to find y values
$y = {\left(- 2\right)}^{3} - 12 \left(- 2\right) - 12$
$y = 4$, $\left(- 2 , 4\right)$
$y = {\left(2\right)}^{3} - 12 \left(2\right) - 12$
$y = - 28$, $\left(2 , - 28\right)$

Using $y ' ' = 6 x$
Sub x values into $y ' ' = 6 x$
$6 \left(- 2\right) = - 12$
$x < 0$
$\therefore$ $\left(2 , - 28\right)$ is a maximum point
$6 \left(2\right) = 12$
$x > 0$
$\therefore$ $\left(- 2 , 4\right)$ is a minimum point

When $x = 0$
$y = {0}^{3} - 12 \left(0\right) - 12$
$y = - 12$
$\left(0 , - 12\right)$

Use the information from the first and second derivative to sketch the graph as well as the y intercept.

graph{y=x^3-12x-12 [-10, 10, -5, 5]}