# How do you find the vertical, horizontal or slant asymptotes for f(x) = (2x^2 + x + 2) /( x + 1)?

May 12, 2016

Vertical asymptote at $x = - 1$
No horizontal asymptote
Slant asymptote: $f \left(x\right) = 2 x - 1$

#### Explanation:

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

$f \left(x\right)$ is undefined when $\left(x + 1\right) = 0$
giving us the vertical asymptote of $x = - 1$

${\lim}_{x \rightarrow \infty} f \left(x\right) \rightarrow \infty$
and
${\lim}_{x \rightarrow - \infty} f \left(x\right) \rightarrow - \infty$
so there is no horizontal asymptote.

Since the degree of the numerator is greater than the degree of the denominator,
we can divide the denominator into the numerator to get a slant asysmptote:
$f \left(x\right) = \left(2 {x}^{2} + x + 2\right) \div \left(x + 1\right) = \left(2 x - 1\right) + \frac{3}{x + 1}$
So the slant asymptote is $f \left(x\right) = 2 x - 1$

Slant Asymptote: $y = 2 x - 1$

Vertical Asymptote: $x = - 1$

Horizontal Asymptote: None

#### Explanation:

The given function is $y = \frac{2 {x}^{2} + x + 2}{x + 1}$

To find the slant asymptote, divide numerator by the denominator of the given rational function.
" " " " " "underline(" "2x-1" " " " " ")
$x + 1 \text{|~" } 2 {x}^{2} + x + 2$
" " " " " "underline(2x^2+2x" " " " " ")
$\text{ " " " " " " " " } - x + 2$
$\text{ " " " " " " " " } \underline{- x - 1}$
$\text{ " " " " " " " " " " " " " } + 3$

The result of the division is

$y = 2 x - 1 + \frac{3}{x + 1}$

The whole number part of the quotient which is $2 x - 1$ becomes the right side part of the linear equation

$\textcolor{red}{y = 2 x - 1}$

which is the $\textcolor{red}{\text{Slant Asymptote}}$

To solve for the Vertical Asymptote, use the divisor and equate to zero

$x + 1 = 0$

and the Vertical Asymptote is

$\textcolor{red}{x = - 1 \text{ is the Vertical Asymptote}}$

There is $\textcolor{red}{\text{No Horizontal Asymptote}}$

Kindly see the graph of the function $y = \frac{2 {x}^{2} + x + 2}{x + 1}$ colored blue, the slant asymptote $y = 2 x - 1$ colored red, vertical asymptote $x = - 1$ colored green below.

God bless....I hope the explanation is useful.