# How do you find the important points to graph f(x)= -x^2-4x?

Jan 18, 2018

Check below for detail examination of the function.

#### Explanation:

$f \left(x\right) = - {x}^{2} - 4 x$ ,
${D}_{f} = \mathbb{R}$

$f \left(x\right) = 0 \iff - {x}^{2} - 4 x = 0 \iff {x}^{2} + 4 x = 0 \iff x \left(x + 4\right) = 0 \iff \left(x = 0 , x = - 4\right)$

$f ' \left(x\right) = - 2 x - 4 = - 2 \left(x + 2\right)$

$f ' \left(x\right) = 0$ $\iff x = - 2$

$f ' ' \left(x\right) = - 2 < 0$
$x$$\in$$\mathbb{R}$

• We get these tables for monotony and concavity of $f$:

$f$ has global maximum at ${x}_{0} = - 2$ , $f \left(- 2\right) = 4$

${\lim}_{x \rightarrow - \infty} f \left(x\right) = {\lim}_{x \rightarrow - \infty} \left(- {x}^{2} - 4 x\right) = {\lim}_{x \rightarrow - \infty} \left(- {x}^{2}\right) = - \infty$

$f \left(- 2\right) = 4$

${\lim}_{x \rightarrow + \infty} f \left(x\right) = {\lim}_{x \rightarrow + \infty} \left(- {x}^{2} - 4 x\right) = {\lim}_{x \rightarrow + \infty} \left(- {x}^{2}\right) = - \infty$

• As a result the range of $f$ will be $\left(- \infty , 4\right]$

$f$ is defined in $\mathbb{R}$ so it doesn't have any vertical asymptotes.

${\lim}_{x \rightarrow - \infty} f \frac{x}{x} = {\lim}_{x \rightarrow - \infty} \frac{- {x}^{2} - 4 x}{x} = {\lim}_{x \rightarrow - \infty} \left(- {x}^{2} / x\right) = + \infty$

• As a result $f$ doesn't have any oblique/horizontal asymptotes.