Question

How do you find the horizontal asymptote for `(x-3) / (x+5)`?

Answer 1

`y=1`

Explanation:

There are two ways of solving this.
1. Limits:

`y=lim_(xto+-oo)(ax+b)/(cx+d)=a/c`, therefore horizontal asymptote occurs when `y=1/1=1`

2. Inverse:
Let's take the inverse of `f(x)`, this is because the `x` and `y` asymptotes of `f(x)` will be the `y` and `x` asymptotes for `f^-1(x)`

`x=(y-3)/(y+5)`
`xy+5x=y-3`
`xy-y=-5x-3`
`y(x-1)=-5x-3`
`y=f^-1(x)=-(5x+3)/(x-1)`

The vertical asymptote is the same as the horizontal asymptote of `f(x)`

The vertical asymptote of `f^-1(x)` is `x=1`, therefore the horizontal asymptote of `f(x)` is `y=1`