In Exercises 12–13, use a vertical shift to graph one period of t...

Question

In Exercises 12–13, use a vertical shift to graph one period of the function.y = 2 cos 1/3 x − 2

Accepted Answer

Welcome back everyone. In this problem, we want to apply a vertical translation to plot a single cycle of the function Y equals three multiplied by the cosine of 1/6 of X minus five. And already I have drawn a sketch of our Y and X axis respectively. Now, what do we already know? Well, we know that this is a trigonometric function and recall that generally, every trigonometric function is in the form Y equals a multiplied by that trick function. In this case, the cosine of BX minus C plus D. If we compare our general form to the function, we have notice that A equals three B is the coefficient of X which is 1/6 we don't have any value for C because there's no phase shift and D is our constant, which in this case is negative five. Now these things are important because our amplitude or our trigonometric graph equals A. So in this case, the amplitude would be three next, our period can be found by using B because our period equals two pi divided by B. So in this case, it would have been two pi divided by 1/6 which would become two pi multiplied by six, which is 12 pi and finally, our vertical shift, vertical shift equals D. So in this case, our vertical shift is negative five. So just from our graph alone, now we can tell that for our cosine graph, its amplitude will be three. In other words, we're going to make it three times bigger. Its period will be 12 pi and we're going to shift it down by five units. So let's go ahead and do that. The first thing we're going to do is to sketch our cosine graph on our axis. So that's what my sketch would look like. And P si had to pause the video and draw this sketch because I find it a little difficult sometimes to draw a smooth sketch. So if you're struggling with it, don't worry too, just get some more practice doing it freehand and you'll be great at it. But this is generally what our cosine graph looks like. Now, we know our period is from 0 to 12 pi. So that means our highest points, our maximums are zero and 12 pi, our minimum is six pi and our cosine graph cuts the X axis uh halfway between those points, which would be three pi and nine pi respectively. Now, remember it said we have an amplitude of three. So that means our highest point, our maximum would be three and our minimum would be negative three. So we've shown that it's a cosine graph, we've made its amplitude three and its period 12 pi all we need to do now is do a vertical shift of negative five and make note of the new points. So a vertical shift of negative five means I'm just going to take my cosine graph and shift it down by five units. And when I do that, it would probably be right there, then three minus five, I've shifted it down by five. So no, the maximum would be negative two and the minimum would be five less than negative three, which would be approximately negative eight, not approximately it would be negative eight. Sorry. Now I'm just going to draw a line in between for reference just for me to get a good idea of where my it it caught the X axis before I know that point, it would have gone from zero down by five units. So it would be at negative five. So that means when X is zero, my Y value would be negative too. When X is three pi my Y value would be negative five when X is six pi then negative eight, when X is nine pi again, it would have been negative five. And then when X is 12 pi or I can just make this one. So in excess 12 pi my Y value is negative two. And that would be a sketch of the function Y equals three multiplied by the cosine of 1/6 of X minus five for a single cycle. Thanks for watching everyone. I hope this video helped.

Struggling with Trigonometry?

In Exercises 12–13, use a vertical shift to graph one period of the function.y = 2 cos 1/3 x − 2

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