How do you find the Vertical, Horizontal, and Oblique Asymptote given (x^3+4)/(2x^2+x-1)?

Feb 20, 2017

The vertical asymptotes are $x = \frac{1}{2}$ and $x = - 1$
The oblique asymptote is $y = \frac{x}{2} - \frac{1}{4}$
No horizontal asymptote

Explanation:

Let start by factorising the denominator

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

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

As, we cannot divide by $0$, $x \ne \frac{1}{2}$ and $x \ne - 1$

The vertical asymptotes are $x = \frac{1}{2}$ and $x = - 1$

As the degree of the numerator is $>$ than the degree of the denominator, we have an oblique asymptote.

Let's do a long division

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

$\textcolor{w h i t e}{a a a a}$${x}^{3} + {x}^{2} / 2 - \frac{x}{2}$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaa)|$\frac{1}{2} x - \frac{1}{4}$

$\textcolor{w h i t e}{a a a a a}$$0 - {x}^{2} / 2 + \frac{x}{2}$$\textcolor{w h i t e}{a a a a}$$+ 4$

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

$\textcolor{w h i t e}{a a a a a a a}$$- 0 + \frac{3}{4} x$$\textcolor{w h i t e}{a a a a a}$$+ \frac{15}{4}$

Therefore,

$f \left(x\right) = \frac{x}{2} - \frac{1}{4} + \frac{\frac{3}{4} x + \frac{15}{4}}{2 {x}^{2} + x - 1}$

${\lim}_{x \to - \infty} f \left(x\right) - \left(\frac{x}{2} - \frac{1}{4}\right) = {\lim}_{x \to - \infty} \frac{\frac{3}{4} x}{2 {x}^{2}} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) - \left(\frac{x}{2} - \frac{1}{4}\right) = {\lim}_{x \to + \infty} \frac{\frac{3}{4} x}{2 {x}^{2}} = {0}^{+}$

The oblique asymptote is $y = \frac{x}{2} - \frac{1}{4}$

graph{(y-(x^3+4)/(2x^2+x-1))(y-x/2+1/4)=0 [-18.02, 18.03, -9, 9.02]}