# How do you find the asymptotes for r(x) = (-1) / (x+ 1)^2?

The asymptotes are: $x = - 1$ and $y = 0$

#### Explanation:

From the given: $r \left(x\right) = - \frac{1}{x + 1} ^ 2$

First question: What will make $r \left(x\right) = \infty$?
Answer: when ${\left(x + 1\right)}^{2} = 0$

Therefore , just solve ${\left(x + 1\right)}^{2} = 0$ for $x$:

${\left(x + 1\right)}^{2} = 0$

$x + 1 = 0$
$x = - 1$ Asymptote

Second question: What happens to the value of $r \left(x\right)$ if $x$ is made to approach $\infty$.

if $x = \pm \infty$

$r \left(x\right) = - \frac{1}{\pm \infty + 1} ^ 2$

$r \left(x\right) = 0$ is an asymptote, whether $x = + \infty$ or $x = - \infty$

Have a nice day!!! from the Philippines.