# How do you identify all asymptotes or holes for f(x)=(x^3+3x^2+2x)/(3x^2+15x+12)?

Nov 17, 2016

There is a hole at $x = - 1$
A vertical asymptote is $\left(x = - 4\right)$

The slant asymptote is $y = \frac{x}{3} - \frac{2}{3}$
No horizontal asymptote

#### Explanation:

Let's simplify $f \left(x\right)$

The numerator $= {x}^{3} + 3 {x}^{2} + 2 x = x \left({x}^{2} + 3 x + 2\right)$
$= x \left(x + 1\right) \left(x + 2\right)$

The denominator $= 3 {x}^{2} + 15 x + 12 = 3 \left({x}^{2} + 5 x + 4\right)$
$= 3 \left(x + 1\right) \left(x + 4\right)$

Therefore, $f \left(x\right) = \frac{x \cancel{x + 1} \left(x + 2\right)}{3 \cancel{x + 1} \left(x + 4\right)}$

There is a hole when $x = - 1$

$\therefore f \left(x\right) = \frac{x \left(x + 2\right)}{3 \left(x + 4\right)}$

As we cannot divide by $0$, $x \ne - 4$

So, a vertical asymptote is $\left(x = - 4\right)$

As the degree of the numerator $>$ degree of the denominator, we expect a slant asymptote.

Let $y = a x + b$ be the slant asymptote

$f \left(x\right) = \frac{{x}^{2} + 2 x}{3 x + 12} = \left(a x + b\right) + \frac{c}{3 x + 12}$

$= \frac{\left(a x + b\right) \left(3 x + 12\right) + c}{3 x + 12}$

$\therefore {x}^{2} + 2 x = \left(a x + b\right) \left(3 x + 12\right) + c$

Comparing the coefficients,

of ${x}^{2}$$\implies$, $3 a = 1$, $a = \frac{1}{3}$

of $x ,$$\implies$, $2 = 3 b + 12 a$, $\implies$ $b = - \frac{2}{3}$

and $12 b + c = 0$$\implies$, $c = 8$

The slant asymptote is $y = \frac{x}{3} - \frac{2}{3}$

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{x}{3} = \pm \infty$

Therefore, no horizontal asymtote

graph{(y-(x^2+2x)/(3x+12))(y-x/3+2/3)=0 [-10, 10, -5, 5]}