Hey, everyone. So let's give this problem a try. Here we're asked to graph the function y=3∙sinx+π. Okay. So how can we solve this problem? Well, what I'm first going to do is write out the most general form for the sin function, which is y=a∙sinbx-h+k. So this is our most general form. Now something that I notice is that we don't have a +k out front here, so we can ignore any kind of vertical shift in our graph. But we do have an a value, and our a value or amplitude is equal to 3. Now what I also notice is for bx-h, what we have in the parenthesis, we don't have a specific b value out front here, so we could just say that b is equal to 1. Right? And you could also say that, like, k is equal to 0, for example, but that's not really necessary for this problem. But if we don't see a number in front of the x, we can just say that number is 1 because one times x would just be x. Now I also notice that we have a π here, but notice that we have a plus π. And the general form of this equation asks for a minus sign. So what we're going to need to do is find a way to get a minus sign in this equation. Well, the way that we can do that is by changing this plus to minus a negative. So we're going to have that y=3∙sinx-−π. And this is a perfectly valid way to write this because these negatives would normally just cancel. So what that does is shows us what our h value is: our h value is negative π. So this is all the details that we have for this trig function. So what we can do is we can try to graph this. Well, since we're dealing with a sine function, and what I'm first going to do to graph this is actually, I'm going to ignore the h value that we have in here. So I'm going to see what I can do to just draw a graph for y=3∙sinx. And on our graph over here, that's going to be a standard sine graph with an amplitude that peaks at 3 on the y-axis and valleys down at negative 3 on the y-axis. So we'll start here at the center like we do for a sine graph, and we're going to reach a peak at π/2, which will be up here at a y value of 3. We're going to cross through π, and then we're going to reach a valley at 3π/2, which will be at a y value of negative 3. This is going to be the valley, and then this graph will keep waving. And this graph is going to keep going to the left as well. So we're going to see a valley here at negative π/2, we're then going to cross through negative π, and then we're going to see a peak at negative 3π/2. And then this graph is going to keep waving. So this is what the graph for 3 times the sine of x would look like. Now what I need to do from here is incorporate the phase shift because we can see that because we have this h value here, there's going to be some kind of phase shift. But the way that I can find that phase shift, and I should actually say the h value is negative pi. And the way that I can find that phase shift is by recognizing the phase shift is going to be h over b units. And since I can see we have a negative h value, it's going to be h over b units to the left. So what I can do is take our h value to find h over b our h value is negative π, but I'm just going to write pi here because we already have this going to the left, And that's going to be divided by b, which is 1. And π/1 is just π. So we can see that this graph is going to shift pi units to the left. So this unit over here, π, is going to go all the way over here. So to rewrite this graph we'll start here at the center, but this whole portion of the graph is going to be moved over. So, we're actually going to see a peak at negative π/2, which reaches 3, then we're going to see that this graph crosses through negative π, and we reach a valley down at negative 3π/2, and we keep waving to the left. And then this graph is going to dip down here and valley at π/2, then we're going to cross through π, and then reach a peak as we get to 3π/2. And then this graph is going to keep waving as we go to the right. So this right here is what the graph would look like for 3 times the sine of x plus pi. This is what our graph would look like, and that would be the solution to this problem. So hope you found this video helpful, thanks for watching.