Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)= (3e^(x))/(2-2e^(x))`?

Answer 1

Vertical asymptote is `x=0` and

two horizontal asymptotes `x=0` and `x=-3/2`

Explanation:

Vertical asymptote is given by putting denominator equal to zero.

Hence it is given by `2-2e^x=0` or `2e^x=2`

or `e^x=1` or `x=ln1=0`

For horizontal asymptote, let us find out `Lt_(x->oo)(3e^x)/(2-2e^x)`

Dividing numerator by `e^x` this is equal to

`Lt_(x->oo)3/(2e^-x-2)=-3/2` as `e^-oo=0`

Also `Lt_(x->-oo)(3e^x)/(2-2e^x)=0`

Hence we have two horizontal asymptotes `x=0` and `x=-3/2`

graph{3e^x/(2-2e^x) [-10, 10, -5, 5]}