Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = ((x-3)(9x+4))/(x^2-4)`?

Answer 1

The vertical asymptotes are `x=2` and `x=-2`
The horizontal asymptote is `y=9`

Explanation:

The denominator, `x^2-4=(x+2)(x-2)`
As we cannot divide by zero, the vertical asymptotes are `x=2` and `x=-2`
As the degree of the numerator is the same asthe degree of the denominator, there is no slant asymptote:
`lim_(x->+-oo)f(x)=lim_(x->+-oo)(9x^2)/x^2=9`

`:. y=9` is a horizontal asymptote
graph{(y-((x-3)(9x+4))/((x+2)(x-2)))(y-9)=0 [-41.13, 41.15, -20.55, 20.57]}