How do you find the asymptotes for f(x)=(2x^3+3x^2+x)/(6x^2+x-1)?

Jul 5, 2018

The vertical asymptote is $x = \frac{1}{3}$. No horizontal asymptote. The slant asymptote is $y = \frac{1}{3} x + \frac{4}{9}$

Explanation:

The numerator is

$2 {x}^{3} + 3 {x}^{2} + x = x \left(2 {x}^{2} + 3 x + 1\right)$

$= x \left(2 x + 1\right) \left(x + 1\right)$

The denominator is

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

The function is

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

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

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

As the denominator must de $\ne 0$, therefore

$3 x - 1 \ne 0$, $\implies$, $x \ne \frac{1}{3}$

The vertical asymptote is $x = \frac{1}{3}$

There is no horizontal asymptote, as

${\lim}_{x \to \infty} f \left(x\right) = {\lim}_{x \to \infty} \frac{{x}^{2}}{3 x} = {\lim}_{x \to \infty} \frac{x}{3} = + \infty$

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{{x}^{2}}{3 x} = {\lim}_{x \to - \infty} \frac{x}{3} = - \infty$

As the degree of the numerator is $>$ the degree of the denominator, there is a slant asymptote.

Perform a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} + x$$\textcolor{w h i t e}{a a a a}$$|$$3 x - 1$

$\textcolor{w h i t e}{a a a a}$${x}^{2} - \frac{1}{3} x$$\textcolor{w h i t e}{a a a a}$$|$$\frac{1}{3} x + \frac{4}{9}$

$\textcolor{w h i t e}{a a a a a}$$0 + \frac{4}{3} x$

$\textcolor{w h i t e}{a a a a a a a}$$+ \frac{4}{3} x + 0$

$\textcolor{w h i t e}{a a a a a a a}$$+ \frac{4}{3} x - \frac{4}{9}$

$\textcolor{w h i t e}{a a a a a a a a a}$$+ 0 + \frac{4}{9}$

Therefore,

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

${\lim}_{x \to + \infty} \left(f \left(x\right) - \left(\frac{1}{3} x + \frac{4}{9}\right)\right) = {\lim}_{x \to + \infty} \frac{\frac{4}{9}}{3 x - 1} = {0}^{+}$

${\lim}_{x \to - \infty} \left(f \left(x\right) - \left(\frac{1}{3} x + \frac{4}{9}\right)\right) = {\lim}_{x \to - \infty} \frac{\frac{4}{9}}{3 x - 1} = {0}^{-}$

The slant asymptote is

$y = \frac{1}{3} x + \frac{4}{9}$

graph{(y-(x(x+1))/(3x-1))(y-1/3x-4/9)=0 [-4.382, 4.386, -2.19, 2.193]}