Question

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

Answer 1

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

Explanation:

`y=2/(x+1)-5`

`y` is defined for all real x except where `x=-1` because `2/(x+1)` is undefined at `x=-1`

N.B. This can be written as: `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]}