# What are the asymptotes for y=2/(x+1)-5 and how do you graph the function?

Jan 25, 2018

$y$ has a vertical asymptote at $x = - 1$ and a horizontal asymptote at $y = - 5$ See graph below

#### Explanation:

$y = \frac{2}{x + 1} - 5$

$y$ is defined for all real x except where $x = - 1$ because $\frac{2}{x + 1}$ is undefined at $x = - 1$

N.B. This can be written as: $y$ is defined $\forall x \in \mathbb{R} : x \ne - 1$

Let's consider what happens to $y$ as $x$ approaches $- 1$ from below and from above.

${\lim}_{x \to - {1}^{-}} \frac{2}{x + 1} - 5 = - \infty$

and

${\lim}_{x \to - {1}^{+}} \frac{2}{x + 1} - 5 = + \infty$

Hence, $y$ has a vertical asymptote at $x = - 1$

Now let's see what happens as $x \to \pm \infty$

${\lim}_{x \to + \infty} \frac{2}{x + 1} - 5 = 0 - 5 = - 5$

and

${\lim}_{x \to - \infty} \frac{2}{x + 1} - 5 = 0 - 5 = - 5$

Hence, $y$ has a horizontal asymptote at $y = - 5$

$y$ is a rectangular hyperbola with "parent" graph $\frac{2}{x}$, shifted 1 unit negative on the $x -$axis and 5 units negative on the $y -$axis.

To find the intercepts:

$y \left(0\right) = \frac{2}{1} - 5 \to \left(0 , - 3\right)$ is the $y -$intercept.

$\frac{2}{x + 1} - 5 = 0 \to 2 - 5 \left(x + 1\right) = 0$

$- 5 x = 3 \to \left(- 0.6 , 0\right)$ is the $x -$intercept.

The graph of $y$ is shown below.

graph{2/(x+1)-5 [-20.27, 20.29, -10.13, 10.14]}