How do you find the asymptote(s) or hole(s) of #f(x) = (x(x^2-4))/((x^2-6)(x-2))#?

1 Answer
Mar 2, 2017

Answer:

Hole at #x=2#, vertical asymptotes at #x=-sqrt6# and #x=sqrt6# and a horizontal asymptote at #y=1#.

Explanation:

As #f(x)=(x(x^2-4))/((x^2-6)(x-2))#.

= #(x(x+2)(x-2))/((x+sqrt6)(x-sqrt6)(x-2))#

= #(x(x+2))/((x+sqrt6)(x-sqrt6))#

Observe that although #(x-2)# has cancelled, we cannot have #x-2=0# as then #f(x)# becomes undefined.

Hence, we have a hole at #x=2#.

As regards asymptotes, #f(x)=(x(x+2))/((x+sqrt6)(x-sqrt6))#

and as #x->+-sqrt6#, #f(x)->+-oo# and as such we have vertical asymptotes at #x=-sqrt6# and #x=sqrt6#.

Further, as #f(x)=(x(x^2-4))/((x^2-6)(x-2))#

= #(x(x+2))/(x^2-6)=(x^2+2x)/(x^2-6)#

= #(x^2-6+2x+6)/(x^2-6)=1+(2x+6)/(x^2-6)=1+(2/x+6/x^2)/(1-6/x^2)#

and as #x->oo#, #f(x)->1#

and hevce, we have a horizontal asymptote at #y=1#.

graph{(x(x+2))/((x+sqrt6)(x-sqrt6)) [-10, 10, -5, 5]}