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

How do you graph $f\left(x\right)=\frac{{x}^{3}-x}{{x}^{3}+2{x}^{2}-3x}$ using holes, vertical and horizontal asymptotes, x and y intercepts?
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scrapbymarieix
You need to first factor to see if you can eliminate anything (this is when holes will occur).

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

There will be two holes: at x=0 and x=1. There will be vertical asymptotes at x=0, x=−3 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 requiblack 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]}