Question

How do you graph `f(x)=(x^3-x)/(x^3+2x^2-3x)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

You need to first factor to see if you can eliminate anything (this is when holes will occur).

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

`f(x) = (x(x + 1)(x - 1))/(x(x + 3)(x - 1))`

`f(x) = (x + 1)/(x + 3)`

There will be two holes: at `x = 0` and `x = 1`. There will be vertical asymptotes at `x = 0, x = -3 and x = 1` (since this is what makes the denominator `0` and hence undefined). However, the supposed vertical asymptote at `x = 1` and `x = 0` is in fact a hole.

The exact coordinates of the holes can be obtained by substituting `x = a` into the simplified function.

Hence, the holes will be at `(0, 1/3)` and `(1, 1/2)`.

For this function, there will be a horizontal asymptote at the ratio between the coefficents of the terms with highest degree in the numerator and denominator.

The horizontal asymptote is given by `y = 1/1 = 1`.

As for intercepts, set the function to `0` and solve.

y intercept:
there are none, because both are eliminated when factoring (even though it does appear that there is a y-intercept on the graph, this is in fact a hole.

x-intercept:

You will find there is an x-intercept at `x= -1`.

The last thing that is required to graph a rational function like this is end behavior. This can be found by picking a few numbers close to the asymptotes and checking their trend. For example, you can pick `x = -3.5` and `x = -3.001`, and on the other side you can pick `x= -2.999` and `x = -2.5`.

Doing this for all the vertical and horizontal asymptotes, you should get the following graph.
graph{(x^3 - x)/(x^3 + 2x^2 - 3x) [-10, 10, -5, 5]}

Hopefully this helps!