# How do you find vertical, horizontal and oblique asymptotes for (3x^2 + 4)/(x+1)?

Nov 7, 2016

The vertical asymptote is $x = - 1$
The oblique asymptote is $y = 3 x - 3$

#### Explanation:

As you canot divide by $0$, the vertical asymptote is $x = - 1$
As the degree of the numerator is $>$ than the degree of the dedenominator, so we expect an oblique asymptote:
Therefore we do a lond division
$3 {x}^{2}$$\textcolor{w h i t e}{a a a a a a}$$+ 4$∣$x + 1$
$3 {x}^{2} + 3 x$$\textcolor{w h i t e}{a a a a}$3x-3
$\textcolor{w h i t e}{a a}$$0 + 3 x + 4$$\textcolor{w h i t e}{a}$
$\textcolor{w h i t e}{a a a a a}$$+ 0 - 3$$\textcolor{w h i t e}{a}$
$\textcolor{w h i t e}{a a a a a a a a a}$$+ 7$$\textcolor{w h i t e}{a}$

So, $\frac{3 {x}^{2} + 4}{x + 1} = 3 x - 3 + \frac{7}{x + 1}$
So the oblique asymptote is $y = 3 x - 3$
${\lim}_{n \rightarrow \pm \infty} \frac{3 {x}^{2} + 4}{x + 1} = {\lim}_{n \rightarrow \pm \infty} 3 x = \pm \infty$
So there is no horizontal asymptote
graph{(y-(3x^2)/(x+1))(y-3x+3)=0 [-43.83, 38.34, -24.63, 16.5]}