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# Graphing the Transformation y = a f(x) + k - 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. This time y equals 8 times 2 to the x minus 5. The first thing we need to recognize is what parent function is being transformed. In this case, it's f(x) equals 2 to the x. What does the graph of that look like? Recall that it looks something like this, sort of an upward sweeping curve, passes through (0, 1) and (1, 2). And as it moves of to the left, it gets closer and closer to the x axis, which is its horizontal asymptote. The x axis also has the equation y equals 0. So we have to remember that y equals 0 is a horizontal asymptote for our parent function, and that means that our transformed graph will have an asymptote as well.

Let’s start plotting some points. Now again, key points for 2 to the x are (0, 1) so when x is 0, y is 1. (1, 2), and we can also use (-1, 1/2). We should keep in mind the horizontal asymptote y equals 0, because the transformations will affect that as well. So what are the transformations? We have 8 times 2 to the x minus 5. Now this function tells me I’ve got to take the 2 to the x values, these values, multiply them by and subtract 5. I don’t really need to do anything to the x so I can just move those over.

Multiply these by 8 and subtract 5. ½ times 8 is 4, minus 5 is -1. 1 times 8 is 8 minus 5 is 3. 2 times 8 is 16, minus 5 is 11. And what happens to y equals 0? It’s also a y value so we could also apply the transformation to it. Multiply by 8 and you get 0 and subtract 5 you get y equals -5. So the new horizontal asymptote is y equals 5. I can plot that now. So -5 is between -4 and 6, so I’ll put it right there. This is my horizontal asymptote.

And then I’m going to plot the rest of my points. I’ve got (-1, -1), so that’s going to be, -1’s here, so that’s one point. I have (0, 3). This is 2, that’s 3. And then I’ve got (1, 11). So this is increasing really fast. That’s 4, that’s 6, 8, 11 is going to be way up there. I think maybe I should plot some points a little further to the left. Let’s go back to our table here.

Let’s go to the left of -1 and try -2 2 to the -2 is ¼. Remember the s value is just translated over, but the y values I’ve got to multiply by 8 and subtract 5. So ¼ times 8 is 2, minus 5 is -3. That’s -2 and that’s -3, so here. I think I’m ready to graph this now. Remembering that it’s got this horizontal asymptote, it’s going to come in, come down kind of like this and approach the asymptote. And then it’s going very steeply up at this point. So that’s my final graph of y equals 8 times 2 to the x minus 5.

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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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