# How do you find the vertical, horizontal or slant asymptotes for f(x)=(2x^3+3x^2+x)/(6x^2+x-1)?

Oct 28, 2016

The vertical asymptote is $x = \frac{1}{3}$
The slant asymptote is $y = \frac{x}{3} + \frac{4}{9}$

#### Explanation:

Let's start by fatorising the numerator and the denominator
$2 {x}^{3} + 3 {x}^{2} + x = x \left(2 {x}^{2} + 3 x + 1\right) = x \left(2 x + 1\right) \left(x + 1\right)$
and $6 {x}^{2} + x - 1 = \left(3 x - 1\right) \left(2 x + 1\right)$

So $f \left(x\right) = \frac{2 {x}^{3} + 3 {x}^{2} + x}{6 {x}^{2} + x - 1} = \frac{x \cancel{2 x + 1} \left(x + 1\right)}{\left(3 x - 1\right) \cancel{2 x + 1}} = \frac{x \left(x + 1\right)}{\left(3 x - 1\right)}$

As we cannot divide by $0$, so the vertical asymptote is $x = \frac{1}{3}$

As the degree of the numerator is greater than the degree of the denominator, we would expect a slant asymptote. So we make a long division.
${x}^{2} + x$$\textcolor{w h i t e}{a a a a}$∣$3 x - 1$
${x}^{2} - \frac{x}{3}$$\textcolor{w h i t e}{a a a}$∣$\frac{x}{3} + \frac{4}{9}$
$0 + \frac{4 x}{3}$
$\textcolor{w h i t e}{a a a a}$$\frac{4 x}{3} - \frac{4}{9}$
$\textcolor{w h i t e}{a a a a a a}$$0 + \frac{4}{9}$

So $f \left(x\right) = \frac{x}{3} + \frac{4}{9} + \frac{\frac{4}{9}}{3 x - 1}$
The slant asymptote is $y = \frac{x}{3} + \frac{4}{9}$
There is no horizontal asymptote as limit of $f \left(x\right)$ as $x \to \pm \infty$ is $\pm \infty$
graph{(x^2+x)/(3x-1) [-7.02, 7.024, -3.51, 3.51]}