How do you find the asymptotes for (2x^2 + x + 2) / (x + 1)?

Jan 10, 2017

The vertical asymptote is $x = - 1$
The slant asymptote is $y = 2 x - 1$
No horizontal asymptote

Explanation:

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

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1\right\}$

As you cannot divide by $0$, $x \ne - 1$

The vertical asymptote is $x = - 1$

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

Let's do a long division

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

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

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

$\textcolor{w h i t e}{a a a a a a a}$$- x - 1$

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

Therefore,

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

So,

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

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

The slant asymptote is $y = 2 x - 1$

graph{(y-(2x^2+x+2)/(x+1))(y-2x+1)(y-50x-50)=0 [-25.64, 25.67, -12.83, 12.84]}