Transforming the Tangent Graph - Problem 3 4,244 views
We’re graphing transformations of the tangent function and I’ve got a really tough example here, y equals negative point 2 tangent of 2x minus pi over 4. This is going to involve some kind of horizontal stretch or compression and a shift. Well it doesn’t matter because my method of substituting theta for 2x minus pi over 4 will allow me to figure out what the x coordinates need to be without worrying about transformations at all. I’ll write up here theta equals 2x minus pi over 4.
Here as usual I’ve got my table of key points and here I want to make a table of points for my graph. Minus point 2 tangent 2x minus pi, that’s an x, minus pi over 4.
Let’s figure out what the x values are going to be. The x values come from this transformation so first I have to add pi over 4 to both sides and then you divide by 2. Theta over 2 plus pi over 8 that’s x, so in each case I need to take these values divide them by 2 and then add pi over 8. Let’s see what we get. Minus pi over 2 divided by 2 is minus pi over 4, plus pi over 8, minus pi over 8. Minus pi over 4 divide by 2 is negative pi over 8 plus pi over 8 is zero. And these keep advancing by pi over 8.
Then we have to do the y values, now the y values are easy. All you have to do is take the tangent value and multiply it by negative point 2. Undefined times negative point 2 is still undefined, negative 1 times negative point 2, point 2, zero, negative point 2 and undefined. Now we have a really nice table of data for one period of my graph.
First thing I want to do is graph the asymptotes which correspond to these points here. Now we’ve got a shift here so I have to be a little bit careful. Negative pi over 8 and 3 pi over 8. Let me call this negative pi over 4 here. I’m going to call this positive pi over 4, maybe I should fill some of these up. Pi over 2, 3 pi over 4 and so on, pi, negative pi over 2. Alright, the first asymptote goes at negative pi over 8 which is right here, right in between zero and negative pi over 4, so let me draw that in and the next asymptote goes at 3 pi over 8 which is halfway between pi over 4 and pi over 2, right there.
It may be a little hard to tell but this distance is exactly pi over 2, that’s my period. If I just keep advancing that part forward, I’ll get the next asymptote, that far backwards the asymptote so I can fill those in without even calculating the values. Of course this assumes that you’re drawing a really careful graph and that your spacing is accurate and consistent. It always pays to be careful when you’re drawing your graph.
The asymptotes are done, now let me plot my key points. So I have 0, pi over 8, pi over 4, point 2, zero, negative point 2. 0, pi over 8, pi over 4. We’ll make this point 2. I’ll make this zero and this negative point 2. So I’ll write point 2 here. Here’s my graph of the first period. And then just duplicate these three points, really easy to duplicate because they line up. That’s the second period and back here I’ll do the third one. Just duplicating these three over here and we’re just about done. There you go. So that’s a graph of three periods of y equals minus point 2 tangent of 2x minus pi over 4.
That was a pretty tricky example because it involved not only horizontal stretching, vertical stretching but also horizontal shift, but we did it all without really thinking too much about transformations. There is the result.