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# Graphing Rational Functions, n less than m - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We’re graphing rational functions starting out with the case, n is less than m and the degree of the numerator is less than m the degree of the denominator. Now here, we have y equals 5 over 4 minus x². The first thing I like to do when I’m graphing rational functions, is plot the x intercepts and the asymptotes. Now x intercepts come from the zeros of the function, which come from the zeros of the numerator. Now this numerator is never going to equal zero, so there are no x intercepts.

What about asymptotes? In this case, the vertical asymptotes will come from zeros in the denominator. The denominator equals zero when 4 minus x² equals zero, x equals 2 or x equals -2. The horizontal asymptotes, because the degree of the denominator is bigger than the degree of the numerator, y equals zero is automatically the horizontal asymptote. I want to make note of that on my graph. Y equals zero is my horizontal asymptote and I’ll draw my vertical asymptotes; x equals negative 2 and x equals 2.

What I usually like to do next is plot a few points. I’ll plot at least one point in each these regions. Let’s start with a point in this left region, -3. I have 5 over 4 minus -3². 4 minus 9, that’s 5 over -5, which is -1. So at -3, we have -1. That’s going to be a point. Let’s try, how about at zero, what happens? At zero we have 5 over 4 minus zero, 5 over 4. 5/4, is about here. And, you may recognize that this is an even function. If you replace x by negative x, you get this exact same expression, and that means that there is going to be symmetry at the y axis. This point will have a reflection over here. There’ll be a point that’s the mirror image of this one across the y axis. I don’t really need to plot 3, I know I’m going to get -1.

How about x equals 1, let’s see what happens in here a little bit more. X equals 1, I get 5 over 4 minus 1², 4 minus 1, 5 over 3. At x equals 1, I get 5/3, which is 1 and 2/3, about here and then again, using symmetry when I have a point here. And keeping in mind the asymptotic behavior, you know that the graph is going to look something like this. It will be symmetrical about the y axis, and, keeping in mind that y equals zero is an asymptote, the function’s probably going to look like this and like this. And it will be symmetric.

That’s it, that’s my graph of y equals 5 over 4 minus x². Now just remember the steps.

First plot x intercepts and asymptotes, then make a table of values. I would say at least one value per region, the regions that the asymptotes divide the plane into. That’s it, that’s your graph.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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