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# Graphing the Transformation y = f(x - h) - 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.

Let's graph a transformation. We have y equals -2/3, the absolute value of x plus 6 plus 4. Now first thing I want to do is identify what parent function is being transformed. It's the absolute value function. So let's start by thinking about y equals the absolute value of x.

Now you recall that the graph of the absolute value function looks something like this. It's got a little corner at the bottom. So we have to think about that when we're graphing our transformed function.

First let's take down the key points of the graph of this function. I'm going to change the variable, and write u, and absolute value of u. So just remember whatever number you plug in -1, 0 or 1, absolute value changes the sign to positive. So this is 1, 0, 1. These are some key points, and this point is particularly important because 0,0 is the location of the turning point of the graph. The vertex. So I'll have to pay close attention to what happens to that during the transformation process.

Now what I'm going to do here is I'm going to make a substitution. I'm going to call this u. U equals x plus 6. If I subtract 6 from both sides, I get u minus 6 equals x. That tells me that to get my x values for this table; I'm making a table now for this function, the transformed function, I need to take the u values from this table, and subtract 6. So I take these guys, and subtract 6. I get -1 minus 6, -7. 0 minus 6, -6. 1 minus 6, -5.

To get the y values from my table, I remember that I have to multiply the absolute value by -2/3. Then I have to add 4. This is the absolute value of u whose values I have right here. So just multiply these by -2/3, and add 4. So 1 times -2/3 is -2/3, plus 4 is 31/3. 0 times -2/3 is 0 plus 4 is 4. This is going to give me the same value as this one did 1 times -2/3 plus 4, 31/3. Those are not great values. So maybe what I should do is think about going backwards a little bit. Let's plug in -3, and 3. I'll get 3, and 3 for the absolute values. Then when I transform the x values, I have to subtract 6, so I get -3 minus 6, -9. -3 minus 6, -3.

To get the y values again I multiply by -2/3, and that's why I picked the 3's. I'll get a nice integer value here. -2/3 times 3 is going to be -2 plus 4 is 2. -2/3 times 3 same thing. You're going to get 2.

Let's plot these points, and keeping in mind that 0, 0 became this point -6, 4. That's going to be my new vertex. -6 well I'll plot that first. Here is -6 and here is 4. Since the graph is going to be basically emanating from this vertex, I only need to plot the two extreme points -9, 2. So let's see this is -8, -9 is here 2 is here, and -3, 2. That's enough to give me a graph of an absolute value function. It's just two lines that are joined at the vertex. That's it. You might want to give the location of the vertex, -6,4, but that's your graph of the transformed absolute value function.

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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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