What are the asymptotes for $y = \frac{2}{x + 1} - 5$ $y=2/(x+1)-5$ and how do you graph the function?

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Answer 1 · 2018-01-25T01:58:52

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

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

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

N.B. This can be written as: $y$ $y$ is defined $forall x in RR: x!=-1$

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

$lim_(x->-1^-) 2/(x+1)-5=-oo$

and

$lim_(x->-1^+) 2/(x+1)-5=+oo$

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

Now let's see what happens as $x-> +-oo$

$lim_(x->+oo) 2/(x+1)-5 =0-5 =-5$

and

$lim_(x->-oo) 2/(x+1)-5 =0-5 =-5$

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

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

To find the intercepts:

$y(0) = 2/1-5 -> (0,-3)$ is the $y-$ intercept.

$2/(x+1)-5 =0 -> 2-5(x+1)=0$

$-5x=3 -> (-0.6,0)$ is the $x-$ intercept.

The graph of $y$ is shown below.

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

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