How do you find the asymptotes for #g(x) = (x^4 -81) /(x^3+3x^2 - x - 3)#?

1 Answer
Feb 22, 2018

Vertical asymptotes at #x=-1#and #x=1# and oblique or slanting asymptote at #y=x#. Wealso have a hole at #x=-3#.

Explanation:

Let us first factorize the numerator #x^4-81# and denominator #x^3+3x^2-x-3#

#x^4-81=(x^2)^2-9^2=(x^2+9)(x^2-9)=(x^2+9)(x+3)(x-3)#

and #x^3+3x^2-x-3=x^2(x+3)-1(x+3)#

= #(x+3)(x^2-1)=(x+3)(x-1)(x+1)#

Hence #g(x)=((x^2+9)(x+3)(x-3))/((x+3)(x-1)(x+1))#

Observe that when we have #x+3=0# or #x-1=0# or #x+1=0#, #g(x)# is not defined. However, as #x+3# appears both in numerator and denominator, this cancels out and we can arrive at limiting value of #f(x)# when #x+3=0# or #x=-3#.

Hence, we have a hole at #x=-3#.

Further, when #x->1# or #x->-1# #g(x)->+-oo# depending on whether we approach these values from left or right.

Hence, we have vertical asymptotes at #x=-1#and #x=1#.

Further as degree of numerator is just one more that of denominator, we should have an oblique or slanting asymptote.

As #y=g(x)=(x^4-1)/(x^3+3x^2-x-3)=(x-1/x^3)/(1+3/x-1/x^2-3/x^3)#

and as #x->+-oo#, #y->x#

Hence, we have an oblique or slanting asymptote at #y=x#

graph{(x^4-81)/(x^3+3x^2-x-3) [-84, 76, -33, 47]}