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 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.
