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

##### 2 Answers
Aug 24, 2015

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.

Aug 27, 2015

$f \left(x\right) = \frac{{x}^{2} + 4 x + 3}{{x}^{2} - 9}$

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

It also has a removable singularity at $\left(- 3 , \frac{1}{3}\right)$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + 4 x + 3}{{x}^{2} - 9} = \frac{\left(x + 1\right) \left(x + 3\right)}{\left(x - 3\right) \left(x + 3\right)} = \frac{x + 1}{x - 3}$

with exclusion $x \ne - 3$

Then

$\frac{x + 1}{x - 3} = \frac{x - 3 + 4}{x - 3} = 1 + \frac{4}{x - 3}$

As $x \to \pm \infty$ we have $\frac{4}{x - 3} \to 0$, so $1 + \frac{4}{x - 3} \to 1$.

That is, $f \left(x\right)$ has a horizontal asymptote $y = 1$

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

$f \left(x\right)$ has a removable singularity (not an asymptote) where $x = - 3$, since

${\lim}_{x \to - 3} f \left(x\right) = {\lim}_{x \to - 3} \frac{x + 1}{x - 3} = \frac{1}{3}$

but $f \left(- 3\right)$ is undefined as both the numerator and denominator are $0$.

Here's a graph of $f \left(x\right)$ 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 $f \left(x\right)$.)