Question

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

Answer 1

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]}