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

Jun 3, 2016

Vertical asymptotes at ${x}_{v} = \left\{2 , - 2\right\}$
Slant asymptote $y = x$

#### 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 ${p}_{m} \left({x}_{v}\right) = 0$
2) horizontal asymptotes when $n \le m$
3) slant asymptotes when $n = m + 1$
In the present case we have ${x}_{v} = \left\{- 2 , 2\right\}$ 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

$\frac{{p}_{n} \left(x\right)}{{p}_{n - 1} \left(x\right)} = {x}^{3} / \left({x}^{2} - 4\right)$
${p}_{n} \left(x\right) = {p}_{n - 1} \left(x\right) \left(a x + b\right) + {r}_{n - 2} \left(x\right)$
${r}_{n - 2} \left(x\right) = c x + d$
${x}^{3} = \left({x}^{2} - 4\right) \left(a x + b\right) + c x + d$

equating coefficients

{ (4 b - d=0), (4 a - c=0), (-b=0), (1 - a=0) :}

solving for $a , b , c , d$ we have $\left\{a = 1 , b = 0 , c = 4 , d = 0\right\}$
substituting in $y = a x + b$

$y = x$