How do you identify all asymptotes or holes for #y=(x^3-1)/(x^2+2x)#?

1 Answer
Jan 19, 2018

Answer:

There is no hole. Vertical asymptotes are #x-0# and #x=-2# and onlique or slanting asymptote is #y=x#

Explanation:

A funcion of the type #(f(x))/(g(x))# is not defined for values for which #g(x)=0#.

But let #g(x)=a(x-alpha)(x-beta)(x-gamma)# and if any of #(x-alpha),(x-beta)# or #(x-gamma)# - say #(x-beta)# is a factor of #f(x)# too, we can still get the value of function as #(x-beta)# cancels out.

We call a hole at #x=beta# , as on both sides of #x=beta#, #(f(x))/(g(x))# is not only defined, the value of the function #(f(x))/(g(x))# can also be find using limits.

Observe that #y=(x^3-1)/(x^2+2x)# can be written as #((x-1)(x^2+x+1))/(x(x+2)# and their is no common factor between numerator and denominator. As such their is no hole.

Further as we have #x(x+2)# in denominator when #x->0# or #x->-2#, from left or right function, the function #y->+-oo#, hence, we have vertical asymptote at #x=0# and #x=-2#.

We have horizontal asymptote if degrees of #f(x)# and #g(x)# are equal and an obliique or slanting asymptote if degree of #f(x)# is just one more than that of #g(x)#.

Hence here we have an obliique or slanting asymptote. As #y=(x^3-1)/(x^2+2x)=(x-1/x^2)/(1+2/x)# and as #x->oo#, we have #y=x# and we have obliique or slanting asymptote as #y=x#.

graph{(x^3-1)/(x^2+2x) [-41.33, 38.67, -22.32, 17.68]}