# How do you identify all asymptotes or holes for f(x)=(x^2-9)/(3x+3)?

Sep 14, 2016

$\frac{{x}^{2} - 9}{3 x + 3}$ has vertical asymptote $x + 1 = 0$ and oblique asymptote $y = \frac{x}{3} - \frac{1}{3}$

#### Explanation:

Asymptotes of algebraic expressions are easy to find,

First of all vertical asymptotes, if denominator has a zero at a point but numerator does not have, we have an asymptote at that point.

In the given expression, as denominator is $3 x + 3 = 3 \left(x + 1\right)$, we have denominator zero for $x = - 1$, but numerator is not zero for this, hence we have vertical asymptote $x + 1 = 0$.

As regards horizontal asymptote, we have it, if the degree of numerator and denominator is equal, but this is not so. Hence, there is no horizontal asymptote.

If the degree of numerator is just one higher than that of denominator, say they are given by $a {x}^{n}$ and $b {x}^{n - 1}$, than we have an **oblique or slanting asymptote is at y=(ax)/b.

Here the degree of numerator is $2$ and that of denominator is $1$, hence for oblique asymptote, we divide ${x}^{2} - 9$ by $\left(3 x + 3\right)$ and we get $\frac{{x}^{2} - 9}{3 x + 3} = \frac{x}{3} - \frac{1}{3} - \frac{8}{3 x + 3}$. As $x \to \infty$, $y = \frac{x}{3} - \frac{1}{3}$

Hence $\frac{{x}^{2} - 9}{3 x + 3}$ has vertical asymptote $x + 1 = 0$ and oblique asymptote $y = \frac{x}{3} - \frac{1}{3}$

graph{(x^2-9)/(3x+3) [-15, 17, -7.5, 7.5]}