Question

How do you find the asympotes for `f (x)= (5x-1)/(x^2 + 9)`?

Answer 1

This one has only a horizontal asymptote.

Explanation:

The vertical asymptote would happen if the numerator would go to `=0`. Since `x^2` is always non-negative, the numerator will always be at least `9`

The horizontal asymptote can be found by making `x` larger and larger. The `+1` and `-9` then make less and less of a difference and the whole thing will tend to look like:
`(5x)/x^2~~5/x`
As `x` gets larger `5/x` gets smaller, or:

`lim_(x->+-oo) f(x)=0`

In fact this is not a real asymptote, as the value of `f(x)=0` is also reached when `5x-1=0->x=1/5` (see graph)
graph{(5x-1)/(x^2+9) [-10, 10, -5, 5]}