Question

Asymptote problems?

Find all the asymptote of the function `y=(sqrt(x^3+3)-x^3-x^2 )/(x^2-x)`

Answer 1

Please see below.

Explanation:

You know that `x` cannot equal `0` because the denominator is `x(x-1)`, but before drawing that vertical asymptote you should check if it makes the numerator `0`.

In that case there would be a hole in the graph. By doing this you find that the numerator wouldn't equal `0` at `x=0` and so you can draw the vertical asymptote at `x=0`.

Doing this process with `x=1` and you find that there isn't a vertical asymptote but a hole at `x=1` as numerator too is `0`.

There is a slant asymptote `y=-x-1` as

`f(x)=(sqrt(x^3+3)-x^3-x^2 )/(x^2-x)=(sqrt(1/x+3/x^4)-x-1 )/(1-1/x)`

and as `x->oo` `y->-x-1`.

Note that the domain cannot go lower than the cube root of `-3` i.e. it cannot be less than `-1.4422`

graph{(sqrt(x^3+3)-x^3-x^2 )/(x^2-x) [-18.3, 21.7, -11.44, 8.56]}