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

Question

1243 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-12T13:50:32

#3/5x#

#(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#

enter image source here