How do you find the asymptotes for (12x^5 + 18x^2) /( 20x^4 + 9x^2)?

Aug 12, 2016

$\frac{3}{5} x$

Explanation:

$\frac{12 {x}^{5} + 18 {x}^{2}}{20 {x}^{4} + 9 {x}^{2}}$ can be simplified to

$\frac{{x}^{2} \left(12 {x}^{3} + 18\right)}{{x}^{2} \left(20 {x}^{2} + 9\right)} = \frac{12 {x}^{3} + 18}{20 {x}^{2} + 9}$

Now performing the division

$12 {x}^{3} + 18 = \left(20 {x}^{2} + 9\right) \left(a x + b\right) + c x + d$

equating to cero $\forall x$ we obtain the relations

{(18 - 9 b + d=0), (-9 a + c=0), (20 b=0), (12 - 20 a=0) :}

Solving for $a , b , c , d$ we obtain

$\left(a = \frac{3}{5} , b = 0 , c = \frac{27}{5} , d = - 18\right)$

then

$\frac{12 {x}^{3} + 18}{20 {x}^{2} + 9} = \frac{3}{5} x + \frac{\frac{27}{5} x - 18}{20 {x}^{2} + 9}$

so, for big $\left\mid x \right\mid$ we have

$\frac{12 {x}^{3} + 18}{20 {x}^{2} + 9} \approx \frac{3}{5} x$

Note.
We don't have vertical assymptotes because $20 {x}^{2} + 9 > 0 , \forall x \in \mathbb{R}$