Asymptote problems?

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

1 Answer
Apr 16, 2018

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]}