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# Graphing the Transformation y = a f(x) + k - 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.

Let’s do another transformation, this time I’m working with y equals 1/3x² minus 2x plus 4. I want to show you kind of a neat trick because I’m sure you know how to graph a quadratic function. This is a quadratic function. But I’ll show you a trick that allows you to use transformations and it makes the graphing very, very easy.

Let’s first factor the 1/3 out of these two terms. I get 1/3 of x². And if you factor a 1/3 out of -2x you get -6x. I’ll just leave the plus 4 as it is. This is actually a transformation of the form a times f(x) plus k where the parent function being transformed, is x² minus 6x. You might think so what? That’s not really a special function, but it’s very, very easy to graph.

So let’s take a look at the graph of x² minus 6x. Since it's factorable x and x minus 6, its graph is going to look like this; (0, 0) and (6, 0). It’s a parabola and it looks like that. And its vertex will be at x equals 3. I can use all of that and then just transform those points using this transformation multiplied by 1/3 and add ¼ to graph the transformed function. So it’s really easy to graph. Let me plot my key points for x and x times x minus 6. And I’ll use 0, 3 and 6 as my x values.

So again, when you plug in 0 you’re going to get 0. When you plug in 6 you’re going to get 0. What happens when you plug in 3? You get 3 times 3 minus 6, -3, -9. Now our transformations says multiply that functions x times x minus 6 by a third and add 4. It’s 1/3 times x² minus 6x plus 4.

Nothing really happens to the x values, so I can just translate these over. 0, 3 and 6, but these guys, the 0, -9 and 0, I'll multiply them by 1/3 and add 4. So 0 times 1/3 is 0, plus 4 is 4. I can do the same thing here. 0 times 1/3 is 0, times 4 is 4, and then -9 times 1/3 is –3, plus 4 is 1. And now, I have to keep in mind that (3, -9), that was the vertex of this function. So this is a special point. This vertex becomes the new vertex, (3, 1). So when I plot that I have to remember that’s the vertex, so (0, 4), (6, 4), and (3, 1). (0, 4) is right here. If this 4, then this is 6. (6, 4) is right here. And then (3, 1), that’s 3, that’s 1 is right here. This is my graph.

And that was really easy to do. You can do this with any quadratic function which is represented as a transformation of some function that is x times some other factor. So very easy method for graphing any quadratic function that uses simple transformations.

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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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