Question

What are the asymptote(s) and hole(s), if any, of `f(x) =1/(e^x-2)`?

Answer 1

There is a asymptote at `x=ln2~~0.6931`

Explanation:

There are no holes because no factors will cancel out of the numerator and denominator, the asymptote(s) are when we have `1/0`

`e^x-2=0`

`e^x=2`

`ln(e^x)=ln2`

`x=ln2`

graph{1/(e^x-2) [-10, 10, -5, 5]}