# When tutoring math 12 or calculus, you encounter graphs of rational functions.  Let’s look at a couple of features:

Rational function graphs are defined by (and you get marks for)  the locations of the asymptotes (if any), as well as the x and y intercepts (once again, provided they exist) and holes (if any).  Today, we’ll look at two of these features:  horizontal asymptotes and holes.

First, to holes:  consider the following rational function:

f(x)=((x-1)(x+2))/((x+2)(x-3)) eqn 1

You can see that, since (x+2) is both in the top and the bottom, the function simplifies to

f(x)=(x-1)/(x-3) eqn 2

However, when you cancel, you are really dividing.  Since you can’t divide by zero, you can’t cancel x+2 when x=-2.  At x=-2, the equation remains undefined.  Therefore, you will get a hole there.  The graph of eqn 1, above, will follow the graph of eqn 2 identically, except for a hole at x=-2.

Now, to horizontal asymptotes:  you get them when the degree on top matches the degree on the bottom or is less than the degree on the bottom.

Case 1:  the degrees on top and bottom are the same.  Consider

f(x)=(2x^2 – 2x -3)/(x^2 +17x+11)

To get the horizontal asymptote, divide the coefficient of the highest exponent term on top by the coefficient of the highest exponent term on the bottom.  The horizontal asymptote will be y=2/1, or just y=2.

Case 2:  the degree on the bottom is greater than that on top.

Simple:  the horizontal asymptote is y=0.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC