Question

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

Answer 1

`x = 4, x = -1 " and " f(x) = 0`

Explanation:

To find the vertical asymptotes:
`lim_{x rarr a^+} f(x) = pm oo " and " lim_{x rarr a^-} f(x) = pm oo`
To find the horizontal asymptotes:
`lim_{x rarr pm oo} f(x) = b`

`f(x) = (2(x+2))/(x^2-3x-4) = (2(x+2))/((x-4)(x+1))`

In this case the vertical asymptotes are when the denominator of `f(x)` in its simplest form is equal to zero. Hence:
`(x-4)(x+1) = 0`
`x = 4 " or " x = -1`

Note: `lim_{x rarr 4^+-} f(x) = +- oo " and " lim_{x rarr -1^+-} f(x) = ""_+^(-) oo`

To find the horizontal asymptotes:
`lim_(x rarr pm oo) f(x) = lim_(x rarr pm oo) (2(x+2))/(x^2-3x-4) = 0^pm`

Hence the asymptotes for `f(x)` are:
`x = 4, x = -1 " and " f(x) = 0`

graph{(2x+4)/(x^2-3x-4) [-5, 8, -10, 10]}