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

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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 $20 x^2+9 > 0, forall x in RR$

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