Question

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

Answer 1

`3/5x`

Explanation:

`(12x^5 + 18x^2) /( 20x^4 + 9x^2)` can be simplified to

`(x^2(12x^3+18))/(x^2(20 x^2+9)) = (12x^3+18)/(20 x^2+9)`

Now performing the division

`12x^3+18 =(20 x^2+9) (a x + b) + 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

`(a = 3/5, b = 0, c = 27/5, d = -18)`

then

`(12x^3+18)/(20 x^2+9) =3/5x +(27/5x-18)/(20 x^2+9)`

so, for big `abs x` we have

`(12x^3+18)/(20 x^2+9) approx3/5x`

Note.
We don't have vertical assymptotes because `20 x^2+9 > 0, forall x in RR`