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

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!