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

Oct 5, 2016

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

$f \left(x\right) = \frac{x \left({x}^{2} - 1\right)}{x \left({x}^{2} + 2 x - 3\right)}$

$f \left(x\right) = \frac{x \left(x + 1\right) \left(x - 1\right)}{x \left(x + 3\right) \left(x - 1\right)}$

$f \left(x\right) = \frac{x + 1}{x + 3}$

There will be two holes: at $x = 0$ and $x = 1$. There will be vertical asymptotes at $x = 0 , x = - 3 \mathmr{and} x = 1$ (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 $\left(0 , \frac{1}{3}\right)$ and $\left(1 , \frac{1}{2}\right)$.

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 $y = \frac{1}{1} = 1$.

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!