Question

What are the asymptotes for `(x^2+4x+3)/(x^2 - 9)`?

Answer 1

The vertical asymptotes is 3.
The horizontal asymptote is 1.

Explanation:

Vertical asymptotes occur at points where the denominator is zero. By factorizing the denominator as a difference of 2 squares, we get (x+3)(x-3), however, factorizing the numerator as a trinomial yields (x+1)(x+3) and so the (x+3) factors cancel, meaning that x=3 is the only vertical asymptote. x=-3 is a point of discontinuity since the function f is undefined at x=-3. One may se the sequential criterion to prove that f(x) is not continuous at x=-3.
For horizontal asymptotes we have to investigate the limits of the function at + and - infinity. In both cases, since the quadratic terms dominate the numerator and denominator, this has value 1 at infinity and so the horizontal asymptote occurs at y=1.

Answer 2

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

has horizontal asymptote `y = 1` and vertical asymptote `x = 3`.

It also has a removable singularity at `(-3, 1/3)`

Explanation:

`f(x) = (x^2+4x+3)/(x^2-9) = ((x+1)(x+3))/((x-3)(x+3)) = (x+1)/(x-3)`

with exclusion `x != -3`

Then

`(x+1)/(x-3) = (x-3+4)/(x-3) = 1+4/(x-3)`

As `x -> +-oo` we have `4/(x-3) -> 0`, so `1+4/(x-3) -> 1`.

That is, `f(x)` has a horizontal asymptote `y = 1`

`f(x)` has a vertical asymptote at `x = 3` where the denominator is `0` (and the numerator is non-zero).

`f(x)` has a removable singularity (not an asymptote) where `x = -3`, since

`lim_(x->-3) f(x) = lim_(x->-3) (x+1)/(x-3) = 1/3`

but `f(-3)` is undefined as both the numerator and denominator are `0`.

Here's a graph of `f(x)` and `y = 1`.

graph{(y - (x^2+4x+3)/(x^2-9))(y - 1) = 0 [-20, 20, -10, 10]}

(The graphing tool does not seem to like graphing the vertical asymptote at the same time as `f(x)`.)