How do you find vertical, horizontal and oblique asymptotes for {-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}?

Jun 14, 2016

Vertical asymptotes ${x}_{v} = \left\{\frac{1}{3} \left(- 3 - \sqrt{3}\right) , \frac{1}{3} \left(- 3 + \sqrt{3}\right)\right\}$
Slant asymptote $y = 3 - 3 x$.

Explanation:

In a polynomial fraction $f \left(x\right) = \frac{{p}_{n} \left(x\right)}{{p}_{m} \left(x\right)}$ we have:

1) vertical asymptotes for ${x}_{v}$ such that ${p}_{m} \left({x}_{v}\right) = 0$. Here ${x}_{v} = \left\{\frac{1}{3} \left(- 3 - \sqrt{3}\right) , \frac{1}{3} \left(- 3 + \sqrt{3}\right)\right\}$
2) horizontal asymptotes when $n \le m$
3) slant asymptotes when $n = m + 1$
In the present case we dont have vertical asymptotes and $n = m + 1$ with $n = 3$ and $m = 2$

Slant asymptotes are obtained considering (p_n(x))/(p_{n-1}(x)) approx y = a x+b for large values of $\left\mid x \right\mid$

In the present case we have

$\frac{{p}_{3} \left(x\right)}{{p}_{2} \left(x\right)} = \frac{- 9 {x}^{3} - 9 {x}^{2} + 28 x + 9}{3 {x}^{2} + 6 x + 2}$
${p}_{3} \left(x\right) = {p}_{2} \left(x\right) \left(a x + b\right) + {r}_{1} \left(x\right)$
${r}_{1} \left(x\right) = c x + d$
$- 9 {x}^{3} - 9 {x}^{2} + 28 x + 9 = \left(a x + b\right) \left(3 {x}^{2} + 6 x + 2\right) + c x + d$

equating coefficients

{ (9 - 2 b - d=0), (28 - 2 a - 6 b - c=0), (9 + 6 a + 3 b=0), (9 + 3 a=0) :}

solving for $a , b , c , d$ we have $\left\{a = - 3 , b = 3 , c = 16 , d = 3\right\}$
substituting in $y = a x + b$

$y = 3 - 3 x$

Note that

$\frac{{p}_{3} \left(x\right)}{{p}_{2} \left(x\right)} = \left(a x + b\right) + \frac{{r}_{1} \left(x\right)}{{p}_{2} \left(x\right)}$

and as $\left\mid x \right\mid$ increases $\frac{{r}_{1} \left(x\right)}{{p}_{2} \left(x\right)} \to 0$

Attached a figure showing the results. 