Question

How do you identify all horizontal and slant asymptote for `f(x)=4/(x-2)^3`?

Answer 1

The vertical asymptote is `x=2`
The horizontal asymptote is `y=0`

Explanation:

As you cannot divide by `0`, you have a vertical asymptote `x=2`
There is no slant asymptote as the degree of the numerator is smaller than the degree of the denominator.
`lim_(n rarr -oo )f(x)=lim_(narr-oo)4/x^3=0^-`

`lim_(n rarr +oo )f(x)=lim_(narr+oo)4/x^3=0^+`

So there is a horizontal asymptote `y=0`
graph{4/(x-2)^3 [-16.02, 16.01, -8.01, 8.01]}