# How do you find the vertical, horizontal or slant asymptotes for f(x) = (3x^2 + 4)/(x+1)?

Jun 29, 2017

Vertical asymptote at $x = - 1$, Slant asymptote: $y = 3 x - 3$ ,
Horizontal asymptote: Absent

#### Explanation:

$f \left(x\right) = \frac{3 {x}^{2} + 4}{x + 1}$
Vertical asymptote : Denominator is $0 \therefore x + 1 = 0 \mathmr{and} x = - 1$

Vertical asymptote at $x = - 1$

Degree i.e maximum power of $x$ in numerator is $1$ more than denominator, so there is no horizontal asymptote.

Slant asymptote: Degree i.e maximum power of $x$ in numerator is $1$ more than denominator , so there is slant asymptote.

By long division of (f(x) we get dividend as $y = 3 x - 3$ and remainder $7 \therefore y = 3 x - 3$ is the slant asymptote.

graph{(3x^2+4)/(x+1) [-80, 80, -40, 40]} [Ans]