# How do you find the critical points to sketch the graph y=x(x+2)^2?

May 27, 2017

#### Explanation:

A critical number for a function $f$ is a number in the domain of $f$ at which either the derivative of $f$ is $0$ or the derivative fails to exist.

$f \left(x\right) = x {\left(x + 2\right)}^{2}$ has domain $\left(- \infty , \infty\right)$

To find $f ' \left(x\right)$ either expand the polynomial or use the product rule.

$f ' \left(x\right) = {\left(x + 2\right)}^{2} + x 2 \left(x + 2\right)$

$= \left(x + 2\right) \left[\left(x + 2\right) + 2 x\right)$

$= \left(x + 2\right) \left(3 x + 2\right)$

$f ' \left(x\right) = \left(x + 2\right) \left(3 x + 2\right)$

$f ' \left(x\right)$ is defined for all $x$ and $f ' \left(x\right) = 0$ at

$x = - 2$ and at $x = - \frac{2}{3}$, both of which are in the domain of $f$.

The critical numbers are $- 2$ and $- \frac{2}{3}$.

If you have been taught that critical point is the same as critical number , then you are done.

You will answer: the critical points are $- 2$ and $- \frac{2}{3}$.

If you have been taught that a critical point is a point on a graph , then you still need to find the corresponding $y$ values using
$y = f \left(x\right) = x {\left(x + 2\right)}^{2}$

$x = - 2$ $\Rightarrow$ $y = 0$
$x = - \frac{2}{3}$ $\Rightarrow$ $y = - \frac{32}{27}$

So you will answer: the critical points are $\left(- 2 , 0\right)$ and $\left(- \frac{2}{3} , - \frac{32}{27}\right)$