In Exercises 53–60, use a vertical shift to graph one period o...

Question

In Exercises 53–60, use a vertical shift to graph one period of the function. y = sin x + 2

Accepted Answer

Welcome back everyone. In this problem, we want to sketch the graph of the function Y equals the sign of X minus eight for one period on our graph. No, what do we already know? Well, we know that this is a trigonometric function and generally for a trigonometric function, it's written in the form a multiplied by that function. In this case, it's the sign of B X minus C plus D. Now, before we get into all of that one thing that I really love about this graph is that it's pretty straightforward. For example, our amplitude would be one right. Next, we don't have any coefficient of X. So our coefficient of X would also be equal to one. So that means our period will remain the same since it's usually two P, two pi divided by B which in this case is one, our period would just be two pi and D which represents our vertical shift how much the graph goes up or down. In this case would just be negative eight like you can see right there. So this tells us that our equation is just going to be a regular sine graph that we've shifted eight units down. We haven't shifted the phase, we haven't done anything else, done nothing to the period. So let's go ahead and sketch that first, let's draw our sine graph and then let's shift it down by eight units. So of course, it starts at zero. I'm just going to put the sketch here, which would look something like this. OK? And the period, as we said, it would be from 0 to 2 pi. So we're just going to want to show where it reaches the halfway point pi and those points in between. So that would be a half of pie here. And I just can't get that pie. Right. Right. Good. And three halves of pie over here because that's halfway between pie and two pie. Now, since it's a regular sand graph, we also know that the peaks will be the peak, the the highest point would be one unit above and the lowest point would be negative one. So we know that already wherever we go, that will be our peaks. So if we're going to shift it down by eight units, I'm just gonna do this neat little thing here where I can take my curve. OK? And I'm going to shift it down by approximately let's call that eight units because it's a sketch. So you know how that is, I'm also going to draw a ver a horizontal line just as a point of reference for myself just so we can put the points in properly and let me, let me mark through it just to make sure we know what's going on. OK. So no, we've shifted our sine graph eight units down and all we need to do on our sketch is to identify the points. So at this point here, we've shifted it down by eight units. So this point would be zero negative eight up here. Remember our peak is one unit above. So this would have been half of pi negative seven right here. We're back to our baseline. So that would be pi negative seven. This point down here we've gone down by one unit. So no, it would have been three oops that three doesn't look good. Three pi divided by two and negative nine. And then finally, this point right here would also have been two pi negative eight. Oh This should have been negative eight as well. Let me draw that a little better. So now this curve would be the sketch of Y equals sine X minus eight. I hope this helped. Thanks for watching.

In Exercises 53–60, use a vertical shift to graph one period of the function. y = sin x + 2

Key Concepts

Vertical Shift in Trigonometric Functions

Period of the Sine Function

Graphing Sine Functions

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