How do you find all the asymptotes for (2x^3+11x^2+5x-1)/(x^2+6x+5 )?

May 22, 2015

The function $\frac{2 {x}^{3} + 11 {x}^{2} + 5 x - 1}{{x}^{2} + 6 x + 5}$ has two vertical asymptotes $x = - 5$ and $x = - 1$ and an oblique asymptote $y = 2 x - 1$.

First of all, we'll have a look on how to find asymptotes in general:

Let $f$ be a function of $x$ $\left(f \left(x\right) = \textrm{s o m e t h \in g w i t h} x\right)$.

1. Vercital asymptotes
We look for vertical asymptotes in points $c$ that are on the "ends" of domain of $f$.

The line $x = c$ is a vertical asymptote when
${\lim}_{x \to {c}^{+}} f \left(x\right) = \pm \infty$ or ${\lim}_{x \to {c}^{-}} f \left(x\right) = \pm \infty$

2. Horizontal asymptotes
We look for horizontal asymptotes in $\pm \infty$ but only when its sensible, meaning the domain of $f$ "streches" towards $\pm \infty$.

The line $y = d$ is a horizontal asymptote when
$d = {\lim}_{x \to + \infty} f \left(x\right)$ or $d = {\lim}_{x \to - \infty} f \left(x\right)$ is a constant (is not $\pm \infty$).

3. Oblique, or slant, asymptotes
We look for oblique asymptotes in $+ \infty$ or $- \infty$, one at a time.

The line $y = a x + b$ is an oblique asymptote when both
$a = {\lim}_{x \to \pm \infty} f \frac{x}{x}$ and $b = {\lim}_{x \to \pm \infty} \left[f \left(x\right) - a x\right]$ are constants.

In our example $f \left(x\right) = \frac{2 {x}^{3} + 11 {x}^{2} + 5 x - 1}{{x}^{2} + 6 x + 5}$.
The domain is ${D}_{f} = \mathbb{R} \setminus \left\{- 5 , - 1\right\}$.

1.
${\lim}_{x \to - {5}^{\pm}} \frac{2 {x}^{3} + 11 {x}^{2} + 5 x - 1}{{x}^{2} + 6 x + 5} = \pm \infty$
so we have a vertical asymptote $x = - 5$
and
${\lim}_{x \to - {1}^{\pm}} \frac{2 {x}^{3} + 11 {x}^{2} + 5 x - 1}{{x}^{2} + 6 x + 5} = \pm \infty$
so we have a vertical asymptote $x = - 1$

2.
${\lim}_{x \to \pm \infty} \frac{2 {x}^{3} + 11 {x}^{2} + 5 x - 1}{{x}^{2} + 6 x + 5} = \pm \infty$
no horizontal asymptotes

3.
$a = {\lim}_{x \to \pm \infty} \frac{2 {x}^{3} + 11 {x}^{2} + 5 x - 1}{x \left({x}^{2} + 6 x + 5\right)} = 2$
and
$b = {\lim}_{x \to \pm \infty} \left[\frac{2 {x}^{3} + 11 {x}^{2} + 5 x - 1}{{x}^{2} + 6 x + 5} - 2 x\right] = - 1$
so we have an oblique asymptote $y = 2 x - 1$.