Hey everyone. So in this example, we're asked to graph this function, y = 2 \sin(x) - 1 . Now, whenever I'm dealing with these types of problems, I like to start these off by solving first what I'm familiar with and graphing it, and then building off the graph from there. So what I'm going to do in this problem is I'm going to start by graphing y = 2 \sin(x) . So, we're going to ignore this negative one for now, but we'll get to that in a moment. To graph this function, for the sine of x, what we do is we start here at our graph. We reach a peak at an output of 1, then we dip through \pi , and this keeps waving. But we can't actually do that for this graph, because we have 2 here, and 2 is going to change the amplitude. So we're still going to start at an output of 0, but now our peaks are going to reach an output of 2, and our valleys are going to reach an output of negative 2. So our graph is going to look something like this. We'll start here at the center of our graph, then we're going to reach a peak when we get up here to \frac{\pi}{2} . So we reach our peak right about there, and then we're going to dip through \pi on our x axis, then we're going to go to \frac{3\pi}{2} , we're going to reach a valley, and then we're going to keep waving as we go to the right. Now what we can do is extend this graph to the left as well. So we're going to reach a valley at -\frac{\pi}{2} , we're going to reach a peak as we go to -\frac{3\pi}{2} , and then we're gonna keep waving to the left. So this is what our sine graph is going to look like when we have 2 \sin(x) . Now to graph y = 2 \sin(x) - 1 , this minus one is going to cause a vertical shift. Since it's a negative value, we're going to have a vertical shift down by 1 unit. Now when we shift down, all of these points are going to be 1 unit down. So what that means is this peak right here is going to go 1 unit down. This valley is going to go 1 unit down. This peak will go 1 unit down, and this valley will go 1 unit down. And then the center here where we started, this is going to actually start at an output of negative one. So what we can do is we can adjust where the peaks and the valleys are going to be, recognizing that the peaks of this graph are actually going to be at positive one when we shift one unit down, and then the valleys of this graph are going to be down here at negative 3. And go ahead and draw this curve, well we'll start here at negative 1, and then we're going to reach our peak right about there when we get to \frac{\pi}{2} . Then we'll cross back down to negative 1 when we get to \pi on the x axis, and then we're going to reach our valley at \frac{3\pi}{2} , which is going to be at an output of negative 3, and then we're going to keep waving to the right. Now, likewise, to the left, we're going to go down here and reach a valley at -\frac{\pi}{2} on the x axis. We're going to go back up here when we get to negative \pi , and then we're going to reach our peak when we get to -\frac{3\pi}{2} , and we're gonna keep waving as we go to the left. So this right here is what the graph is going to look like for y = 2 \sin(x) - 1 . It's going to be this red graph right here, and that is the solution to this problem. So, I hope you found this video helpful. Thanks for watching.