 In this video, we present the solution to question number 14 for practice exam number two for math 1060, in which case we're asked to graph the function f of x equals 3 plus 2 sine of one half x minus pi over 2. And we want to do this for at least one cycle. You can do more than that if you want to, but we need to do at least one cycle here. And so some things we have to do is we do have to label at least three pictures, three points on the graph. I'm actually going to do five because honestly drawing a sine wave, it's just easier to have five points. So I'm just going to upgrade that. And then we need to indicate all the transformations we've done to y equals sine of x to get here. And that's where we're going to start. And we're going to list these things right here in the blank. But before we do that, I first want to put this into standard form. So this thing should look like f of x equals 3 plus 2 sine of inside of the angle here. This needs to be factored. Whenever there's a coefficient in front of the x, we need a factor away from the x and the constant if it's present there. So factoring out the one half, we're left behind with x minus pi like so. And so in this proper standard form, we can see the transformations more readily here. So what does this plus three here represent the plus three represents that we're going to have a shift up. By three units. So k equals three right here. It represents a shift up in which case, if we just mark this as like one, two, three, four, et cetera on the y axis, that means our midline has been moved up to y equals three. You do not have to include the midline with your graph, but it makes it a lot easier. So I am going to include it. So that's what this plus three is doing for us. What does this negative pi here represent? This is our h value pi. This is going to represent a shift to the right by pi. So instead of starting, say at the origin, as things got shifted up, this is a sine wave. We'd start here. I'm actually going to take a step to the right by pi. But we'll come back to that in just a second. What is this plus two here represent? This is not plus two times two. This is our amplitude a equals two. So the amplitude has been changed. So which means we can go to above the midline and to below the midline. So we can come all the way up to five and all the way down to one. And just couldn't put those marks in the y axis. So I remember that. And then the last thing to remember here is that this value B is one half. That's the coefficient of X right there. That affects the period for which the period is going to be two pi divided by one half. That is four pi. So we need a graph basically from zero to four pi. But JK, right? Because of the shift to the right by pi, I'm actually going to graph this thing from pi to five pi. Looking at my X coordinates here, the X axis, I think the easiest way to accomplish this is be, let's make every tick mark be a pi half. So every two would be pi. So we're going to get pi, two pi, three pi, four pi, and five pi. There's no reflections going on here. And so with this information, I'm now ready to start graphing my sine wave. So I'm going to start on the midline at pi because I took a step to the right. Then sine goes up to its maximum. That's going to happen at two pi. It'll return to the midline at three pi. It'll go to its minimum at four pi. And then it returns to the midline at five pi. And so connecting the dots, we get this picture right here. Let's label them. So we're going to get pi, three. We're going to get two pi, five. We're going to get three pi, three. We're going to get four pi, one. And then we're going to get five pi, three. So again, it said indicate three points. I'm labeling five because that gives us the whole graph right there. So we listed our transformations. We've drawn the graph. We have all the points indicated. And so this didn't finish this question number 14.