# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)=(2x^3+3x^2+x)/(6x^2+x-1)?

Jun 24, 2016

A vertical assymptote at $x = \frac{1}{3}$ and
a slant assymptote given by $y = \frac{x}{3} + \frac{4}{9}$

#### Explanation:

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

We have a vertical assymptote at $x = \frac{1}{3}$ and also we know

$\frac{1}{3} x \left(x + 1\right) = \left(x - \frac{1}{3}\right) \left(a x + b\right) + c$

Equating power coefficients

{ (b/3 - c=0), ( 1/3 + a/3 - b=0), (1/3 - a=0) :}

and solving

$a = \frac{1}{3} , b = \frac{4}{9} , c = \frac{4}{27}$

The assymptote is given by

$y = \frac{x}{3} + \frac{4}{9}$

Attached a figure with vertical and slant assymptotes