Question

How do you find the asymptotes for `(x-4)/(x^2-3x-4)`?

Answer 1

You start by factoring the denominator, which can be written as `(x+1)(x-4)`

Explanation:

Now the vertical discontinuities are clear, because that's when the denominator becomes zero:
`x=-1andx=+4`

If we near `x=4` from both sides, it will be seen that:
`lim_(x->4+) y = lim_(x->4-) y=1/5`
So this is a 'repairable' discontinuity, while `x=-1` is not.

We can now do the following:
`(cancel(x-4))/((x+1)(cancel(x-4))) =1/(x+1)`

The horizontal asymptote is when `x` becomes very large. In this case the whole thing goes to zero.

Answer:
`x=-1`
`y=0`
graph{(x-4)/(x^2-3x-4) [-10, 10, -5, 5]}