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.

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The Transformation y = f(bx) - 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.

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Let’s graph another transformation of the type y equals f(bx). We have g(x) here equals 2x times 2 minus x. I want to graph that function and y equals g(1/2x). First g(x). Now let’s make it u and g of u. I want to plot a few points but let me first recognize that this is going to be a quadratic function. So its graph will be a parabola.

And I can first plot its x intercept 11 intercept at 0 0 and at 2 0. So when u equals 0, y will equal 0. And when u equals 2,y will equal 0. And that suggests that the vertex which is always right half way between the x intercepts will be at 1.

So I plug in 1, I get 2 times 1, 2. So this is my vertex I just want to remember that. Because when I transform this point, I want to be aware that its transformation will also be the vertex of the new graph. So I have 0 0, 2 0 and 1 2, let me plot those really quickly.

And so that’s going to look something like this. The transformation. First of all, y equals g of ½x is 2 times 1/2x times 2 minus ½ x. So that’s what I’m going to graph here; x and g of 1/2 x. So I’ll make a substitution I’ll call ½x u. And if u equals to 1/2x then x equals to u. And that tells me how I’m going to get these x values. I’m going to multiply these u values by 2. So 0 times 2, 0, 1 times 2, 2 and 2 times 2 is 4. Notice this is just g of u. It’s exactly the same as this.

The y values I just copy over. 0 2 0. So now I’ve got a parabola with intercepts at 0 0 and 4 0 and a vertex. Remember this point is a transformation of this at 2 2. So 0 0, 4 0 and 2 2. And that gives me this graph.

And you’ll notice the orange graph is the graph of g(1/2x). 1/2x you might think intuitively it seems like that would be a horizontal compression. But it’s actually a horizontal stretch by a factor of 2. And how’s the 2 related to the ½? It’s the reciprocal.

So remember when you are transforming functions of this type y equals f(bx). Remember that b is greater than 1. You get a horizontal compression, so you need that value to be bigger than 1. And the compression factor will be 1 over b.

However, if it’s like it was in this case and b is between 0 and 1, you get a horizontal stretch. And the stretch factor is also 1 over b. So these are our final graphs, the stretched graph and the original parabola.

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