Question

How do you find the vertical, horizontal or slant asymptotes for `(e^x)/(7+e^x)`?

Answer 1

We do not have any other asymptote, just two horizontal asymptotes, `x=0` and `x=1`

Explanation:

Dividing numerator and denominator by `e^x` in `e^x/(7+e^x)`

we get `1/(7e^(-x)+1)`

As `x->oo`, `e^(-x)->1/e^oo=0`

Hence `e^x/(7+e^x)=1/(7e^(-x)+1)->1/(0+1)=1`

Hence the asymptote is `y=1`

When `x->-oo`, `e^x/(7+e^x)=e^(-oo)/(7+e^(-oo))=0/(0+7)=0`

Hence other asymptote is `y=0`

We do not have any other asymptote, just two horizontal asymptotes

graph{e^x/(7+e^x) [-10, 10, -2, 2]}