Question

How do you find the vertical, horizontal or slant asymptotes for `y = (2e^x) / (e^x - 5)`?

Answer 1

We have a vertical asymptote at `x=ln5=1.6094`

and horizontal asymptotes at `y=0` and `y=2`

Explanation:

We have `y=(2e^x)/(e^x-5)`

As `x->-oo`, `y->0/(0-5)=0`

dividing numerator and denominator by `e^x`, we get

`y=2/(1-5/e^x)`

As such, when `x->oo`, `y->2/(1-0)=2`

Hence, horizontal asymptotes are `y=0` and `y=2`

also `ye^x-5y=2e^x` or `ye^x-2e^x=5y`

i.e. `e^x=(5y)/(y-2)` or `x=ln((5y)/(y-2))=ln(5/(1-2/y))`

and as `y->+-oo`, `x->ln5=1.6094`,

we have a vertical asymptote at `x=ln5=1.6094`

graph{(2e^x)/(e^x-5) [-10, 10, -5, 5]}