# Transforming Secant and Cosecant - Problem 3 3,012 views

Let’s graph a transformation of the secant function; y equals -0.5 secant. X over 3 plus pi over 6. Let’s start by recalling the key points of secant. Secant has two vertical asymptotes at negative pi over 2 and pi over 2.

So at these points, secant is undefined. And then half way between a 0, secant is 1. A third away from negative pi over 2 to 0 negative pi over 3, secant is 2. And it’s the same at pi over 3 because secant is an even function.

So what’s our transformed graph going to look like? X over 3 plus pi over 6. Well let’s make the substitution u equals x over 3 plus pi over 6. If we do that, we can solve this for x. I subtract pi over 6 and then multiply everything by 3. I get 3u minus pi over 2 equals x. That means to get my x values all I have to do, is multiply my u values by 3 and subtract pi over 2. So multiply this by 3 and I get -3pi over 2, minus pi over 2 is -4 pi over 2, -2pi. Negative pi over 3 times 3 is negative pi, minus pi over 2 is -3pi over 2.

0 times 3 minus pi over 2, negative pi over 2. Pi over 3 times 3, pi, minus pi over 2, pi over 2. And then pi over 2 times 3 is 3pi over 2, minus pi over 2 is pi.

Now how do I get the y values? Well this is just secant u. It’s exactly the same as this. Only I have to multiply these values by -0.5. So undefined times -0.5, is still undefined here, and here and then 2 times -0.5 is -1. -1 and then 1 times -0.5 is -0.5. So this means I’m going to have vertical asymptotes so s equals -2pi and an x equals pi.

Let me plot those first. So here’s x equals -2pi, here’s x equals pi. And once you have 2 asymptotes graphed, you know that the asymptotes are going to be separated by 3pi, just like these are. And so if I go 3pi to the right I’ll get another asymptote, here at 4pi. If I go 3pi to the left, I’ll get another so I have 1 -5pi. And so on. Let me plot some points.

So we have these 3 points in the middle negative pi over 2, -0.5. Negative pi over 2 is right here between these two vertical asymptotes and I get -0.5 that’s right here. Then add -3pi over 2, I get -1. -3pi over 2 is 1 third the way from -2pi, to here at pi over 2 right here. I get -1 and then at pi over 2, I get -1. This is pi so that’s pi over 2 -1 again.

You will notice my first half period is upside down and that’s because of the vertical reflection. This -0.5 gives me a reflection across the x axis. Now once I have a half period graphed, I can get another half period by taking this flipping it across the x axis and shifting right half period.

Now in this case, it looks like the period is 6pi, so half a period is 3pi. So I take each of these points flip and shift 3pi. So -1/2 becomes positive a half then I shift 3pi to the right. Right there, this point -1 because +1. Shift to the right and I get this point here. And this point flipped up and shifted right here and so I get my second half period.

And once you have two half periods, you can extend this in both directions really easily using periodicity. So this u shape, if I shift it 2 up here to the left I get a point here, a point here another half period of my secant graph. That I can do it over here too. So I get this point shifted 3pi 6 pi to the right becomes this, this one becomes, this and I’ll just do half of the u shape. That looks like 1 and 3/4 periods of my secant graph. So just remember when you got a secant graph, drop graph 1 half period then flip it, shift it to the right, get another half period then use periodicity to fill out your graph to the left and right.