Hey everyone. So here we have an example where we are asked to graph the function y=12tan⁡x−π2. Now to solve this problem, what I'm going to do is I'm first going to graph the function y=12tan⁡x. So I'm going to ignore this π2 for now, but we will get to this later. So let's start with this function.

To find y=12tan⁡x, well, let's just recall what the tangent graph looks like. I know for the tangent graph, we have asymptotes at negative π2, and at π2. Now we do have more asymptotes on this graph, we're going to have other asymptotes at odd multiples of π2, but for now I'm just going to draw these two.

Now as for the curve, I know the curve starts here in the middle, and then it goes up to the right, and then down and to the left. Now another thing that I recall for this tangent is that we have points at π/4 ,1, and then at negative π/4 ,−1. So the graph ends up looking something like this. But notice that we have a one half in front of our tangent. And this one half, as you may recall, changes the amplitude or tallness of our graph. So rather than having these points at 1 and negative one on the y-axis, these are going to be reduced. So we're then going to have a point instead at one-half, and then at negative one-half for these two values of π/4 and negative π/4 respectively.

So that means the graph is going to look a little bit more like this, where we can see that we're a little shorter than we were before. So this right here would be the curve for y=12tan⁡x.

Now our last step for solving this problem is going to be to incorporate this π2. So we need to figure out what y=12tan⁡x−π2 looks like. Well, something that I can see is that we have a π2 here which is going to shift our graph in some sort of way. And to find the shift we'll recall that we can find the h over b ratio, which tells us how much we've shifted to the left or to the right. Now see that we have a minus sign here, which means that this is actually a positive h value. So we have positive π/2 divided by b, which in this case is any number in front of the x, but since there's nothing there we can just write 1. So we're going to have h over b is equal to just π/2, because this one in the denominator is just going to keep this the same. So that means that our graph is going to be shifted π/2 units to the right. So what I can do is take this point right here, and I can shift it π/2 units to the right.