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

Jun 4, 2016

Vertical asymptote at ${x}_{v} = 0$
Slant asymptote $y = 3 x$

#### 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 ${p}_{m} \left({x}_{v}\right) = 0$
2) horizontal asymptotes when $n \le m$
3) slant asymptotes when $n = m + 1$
In the present case we have ${x}_{v} = 0$ and $n = m + 1$ with $n = 4$ and $m = 3$

Slant asymptotes are obtained considering

$\frac{{p}_{n} \left(x\right)}{{p}_{n - 1} \left(x\right)} \approx y = a x + x$

for large values of $\left\mid x \right\mid$.
In the present case we have

$\frac{{p}_{n} \left(x\right)}{{p}_{n - 1} \left(x\right)} = \frac{3 {x}^{4} + 4}{{x}^{3} + 3 x}$

then

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

equating coefficients

{ (4 - e=0), (-3 b - d=0),( -3 a - c=0),( -b=0),(3 - a=0) :}

solving for $a , b , c , d , e$ we have $\left\{a = 3 , b = 0 , c = - 9 , d = 0 , e = 4\right\}$
substituting in $y = a x + b$

$y = 3 x$