How do you find vertical, horizontal and oblique asymptotes for #x^3/(x^2-4)#?

1 Answer
Jun 3, 2016

Answer:

Vertical asymptotes at #x_v = {2,-2}#
Slant asymptote #y = x#

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 = {-2, 2}# and #n = m+1# with #n = 3# and #m = 2#

Slant asymptotes are obtained considering #(p_n(x))/(p_{n-1}(x)) approx y = a x+b# for large values of #abs(x)#

In the present case we have

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

equating coefficients

#{ (4 b - d=0), (4 a - c=0), (-b=0), (1 - a=0) :}#

solving for #a,b,c,d# we have #{a = 1, b = 0, c = 4, d = 0}#
substituting in #y = a x + b#

#y = x #

enter image source here