Question

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

Answer 1

using lim calculus

Explanation:

vertical asymptote:
first find domain: `x^2-9!=0` then `x!=+-3`
then calculate
`lim f(x)`when `x->+3` and it is `24/0 =oo`
but is not the same for `x->-3`
so you have vertical asymptote x=3

horizontal asymptote;
calculate
`lim f(x)` when `x->oo` that's 1
so you have horizontal asymptote for y=1

when you have horizontal asymptote you haven't oblique ones

Answer 2

vertical asymptote x = 3
horizontal asymptote y = 1

Explanation:

The first step is to factorise and simplify f(x).

`f(x)=(cancel((x+3))(x+1))/(cancel((x+3))(x-3))=(x+1)/(x-3)`

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 3 = 0 → x = 3 is the asymptote

Horizontal asymptotes occur as `lim_(xto+-oo),f(x)toc" ( a constant)"`

divide terms on numerator/denominator by x

`(x/x+1/x)/(x/x-3/x)=(1+1/x)/(1-3/x)`

as `xto+-oo,f(x)to(1+0)/(1-0)`

`rArry=1" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ). Hence there are no oblique asymptotes.
graph{(x^2+4x+3)/(x^2-9) [-10, 10, -5, 5]}