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

##### 2 Answers

#### Answer:

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:

has horizontal asymptote

It also has a removable singularity at

#### Explanation:

with exclusion

Then

As

That is,

but

Here's a graph of

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