How do you find vertical, horizontal and oblique asymptotes for #G(x)=(3*x^4 +4)/(x^3+3*x)#?

1 Answer
Jun 4, 2016

Answer:

Vertical asymptote at #x_v = 0#
Slant asymptote #y = 3x#

Explanation:

In a polynomial fraction #f(x) = (p_n(x))/(p_m(x))# we have:

#1)# vertical asymptotes for #x_v# such that #p_m(x_v)=0#
#2)# horizontal asymptotes when #n le m#
#3)# slant asymptotes when #n = m + 1#
In the present case we have #x_v =0# and #n = m+1# with #n = 4# and #m = 3#

Slant asymptotes are obtained considering

#(p_n(x))/(p_{n-1}(x)) approx y = a x+x#

for large values of #abs(x)#.
In the present case we have

#(p_n(x))/(p_{n-1}(x)) = (3x^4+4)/(x^3+3x)#

then

#p_n(x)=p_{n-1}(x)(a x+b)+r_{n-2}(x)#
#r_{n-2}(x)=c x^2 + d x+e#
# (3x^4+4) = (x^2-4)(a x + b) + c x^2 + d x + e#

equating coefficients

#{ (4 - e=0), (-3 b - d=0),( -3 a - c=0),( -b=0),(3 - a=0) :}#

solving for #a,b,c,d,e# we have #{a =3, b = 0, c = -9, d = 0,e=4}#
substituting in #y = a x + b#

#y = 3x #

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