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

1 Answer
Aug 12, 2016

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

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