Question

How do you find all the asymptotes for function `f(x)= (3e^(x))/(2-2e^(x))`?

Answer 1

Vertical asymptote: `x=0`
Horizontal asymptotes:
`y=0`
`y=-3/2`

Explanation:

You start by checking which values of `x` make your denominator equal to zero (you do not want this!).
To avoid zero in the denominator `x` must be different from zero or:
`x!=0` this means that the vertical line of equation `x=0` will be a "forbidden zone", i.e., a vertical asymptote.

To see if we have horizontal asymptotes we check the behaviour of your function for `x` very large or, using the idea of Limit :

1] `x->+oo`
`lim_(x->+oo) (3e^x)/(2-2e^x)=lim_(x->+oo) (3e^x)/(e^x(2/e^x-2))=`
`=lim_(x->+oo) (3cancel(e^x))/(cancel(e^x)(2/e^x-2))=` and when `x->+oo`:
`=lim_(x->+oo) 3/(-2)=-3/2` (where I used the fact that `1/e^oo=1/oo`)
So, the horizontal asymptote will be the horizontal line of equation: `y=-3/2`.

2] `x->-oo`
`lim_(x->-oo) (3e^x)/(2-2e^x)=lim_(x->-oo) (3e^x)/(e^x(2/e^x-2))=`
`=lim_(x->-oo) (3cancel(e^x))/(cancel(e^x)(2/e^x-2))=` and when `x->-oo`:
`=lim_(x->-oo) 3/(2/e^x-2)=0` (where I used the fact that `1/e^-oo=e^oo`).
So, the horizontal asymptote will be the horizontal line of equation: `y=0`.

Graphically:
graph{(3e^x)/(2-2e^x) [-11.25, 11.25, -5.625, 5.625]}