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# Transforming the Cotangent Graph - Problem 3

###### 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.

For graphing transformation as the cotangent function let's try a harder example. Graph y equals 2 cotangent one quarter theta minus 2. Now let’s start by listing some key points for our cotangent function.

I like to graph between 0 and pi, we have pi over 2 pi over 4 and 3 pi over 4. At 0 and pi cotangents undefined pi over 2 it's 0 at pi over 4 it's 1 and at 3 pi over 4 it's -1. So how’s that going to affect the graph of 2 cotangent one quarter theta minus 2?

Well I’ll make the substitution. U equals one quarter theta and that way theta equals 4u, so in order to find my values of theta I have to multiply these values by 4. So 0 times 4, 0, pi over 4 times 4, pi, pi over 2 times 4, 2pi 3pi and 4pi. Now to get my y values I have to take cotangent of u in the values I have here, multiply by 2 and subtract 2.

So 2 times undefined minus 2 it's still going to be undefined same thing here. But 2 times 1 minus 2 is 2 minus 2,0. 0 times 2 minus 2 -2. -1 times 2 -2, minus 2 is -4. So this will give me points that I can use to plot one good period of the cotangent graph.

Let me start by graphing the two vertical asymptotes x equals 0 and x equals 4pi. So x equals 0 is here right along the y axis, and x equals 4 pi is here. And it don’t matter I might as well graph other asymptotes. I know that these first two asymptotes define one period of my cotangent graph so the period is going to be 4 pi. That means I'll have asymptotes every 4pi so one at 8pi and one at -4pi as well. Now let me continue by graphing some key points between 0 and 4pi, these points. Pi, 0, 2pi,-2 and 3pi,-4.

Pi, 0, 2pi,-2 and 3pi,-4 and remember cotangent has a decreasing shape so kind of like this. So if I want to graph another period I just shift these three points over so this point goes over 4pi to here, this one goes to here, this one goes to here. Then I can shift them over this way as well I get one here, one here and one here then here we go.

We’ve got three periods of my cotangent graph notice it shifted down a little bit, the inflection points of the graph are below the x axis they are usually on it, but this is three periods of my function y equals 2 cotangent one quarter theta minus 2.

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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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