# How do you find vertical, horizontal and oblique asymptotes for (x^3 + 1) / (x^2 - 2x + 2)?

May 28, 2017

The oblique asymptote is $y = x + 2$
There is no vertical asymptote
No horizontal asymptote

#### Explanation:

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

The degree of the numerator is greater than the degree of the denominator, so there is an oblique asymptote.

Let's perform a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} - 2 x + 2$$\textcolor{w h i t e}{a a}$$|$${x}^{3} + 0 {x}^{2} + 0 x + 1$$\textcolor{w h i t e}{a a a a}$$|$$x + 2$

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

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$0 + 2 {x}^{2} - 2 x + 1$

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

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a}$$0 + 2 x - 3$

Therefore,

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

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

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

The oblique asymptote is $y = x + 2$

The denominator is

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

$\forall x \in \mathbb{R} , {\left(x - 1\right)}^{2} + 1 > 0$

There is no vertical asymptote
graph{(y-(x^3-1)/(x^2-2x+2))(y-x-2)=0 [-10.97, 11.53, -2.97, 8.28]}