Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `((x-3)(9x+4))/(x^2-4)`?

Answer 1

The vertical asymptotes are `x=2` and `x=-2`
The horizontal asymptote is `y=9`
No slant asymptote

Explanation:

Let `f(x)=((x-3)(9x+4))/(x^2-4)`

Let's factorise the denominator

`x^2-4=(x+2)(x-2)`

As you cannot divide by `0`, `x!=2` and `x!=-2`

The vertical asymptotes are `x=2` and `x=-2`

As the degree of the numerator is `=` the degree of the denominator, there is no slant asymptote

`lim_(x->+-oo)f(x)=lim_(x->+-oo)(9x^2)/x^2=9`

The horizontal asymptote is `y=9`

graph{(y-((x-3)(9x+4)/(x^2-4)))(y-9)=0 [-41.1, 41.1, -20.55, 20.57]}