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# The Transformation y = f(bx) - Problem 2

###### 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 another transformation. I want to graph y equals 4 minus 2 to the ½x plus 1. Now this is a transformation of the parent function; y equals 2 to the x. Now recall that that function looks like this. It’s an upward sweeping curve, it passes through the point 0 1 and 1 2. And it’s got this horizontal asymptote at y equals 0. That means that as we go up to the left the graph gets closer and closer to the line y equals 0. So I have to keep in mind as I’m graphing both as I’m graphing the transformation. But first can we make a table of data for this parent function, and I’ll use the variable u and 2 to the u.

I usually use nice easy values like -1 0 and 1. 2 to the -1 is ½. 2 to the 0 is 1, 2 to the 1 is 2. And I want to record that y equals 0 is the horizontal asymptote. Because I’m going to plot the transformation to the asymptote as well us to the points.

I have x and I’m going to rewrite this as -2 to the ½ x plus 1 plus 4. I’ll put the plus 4 at the end. First of all I, want to rename this part u. So if u equals 1/2x plus 1 what is x equal? I subtract 1 for both sides and I get 1/2x and then I multiply both sides by 2. And I get I’ll write down here x equals 2 u minus 2.

So this is how I’m going to get my x values I’ll take the u values, double them and subtract 2. So if I double this is get -2 and subtract 2 I get -4. If I take this and double it I get 0 subtracting 2 I get -2. If I take 1 and double it I get 2 minus 2 is 0.

Now what about the y values? Well I’ve got 2 to the u, I have to take the opposite of 2 to the u and add 4. So the opposite of a half is -1/2 plus 4 ,3.5. The opposite of 1 is -1 plus 4 is 3. The opposite of 2 is -2 plus 4 is 2.

And let’s not forget the horizontal asymptote. Since these are all y values I can plot the same transformation to this y value. So the opposite of 0 is 0 plus 4 is 4. So y equals 4 will be my new asymptote. And in fact that’s the first thing I want to graph.

I will graph my asymptote at y equals 4. Let me pick another color for that. So this is going to be my horizontal asymptote. And then I’ll fill in my points; -4, 3.5. So here is -4, this is 4 so 3.5 is half way between 3 and 4, right about there, -2, 3, -2 is here and 3 is here. So and 0 2 that’s here. Looks like I can use one more point. So why don’t I use let’s say 2. 2 to the 2 is 4.

Remember I take the u values, I double them and subtract 2. So 2 times 2 is 4, minus 2 is 2. And I take this y value, take its opposite. And add 4 so I get minus 4 plus 4, 0. So 2,0 is another point and that’s this point here. That will give me a pretty good graph. So let me just draw, there it is. Asymptote at all we have this horizontal asymptote y equals 4 and this downward curving graph. And it curves downward because of this minus sign.

That’s a reflection across the x axis. It's been shifted up 4 units. That’s where the asymptote is up at y equals 4 rather than at 0. And also there’s been some horizontal shifting and horizontal stretching as well. So you could see that in the effect of the graph being moved a little bit it’s been made a little bit wider. So this is our final 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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