We're asked to determine theamplitude and the period of y equals negative1/2 cosine of 3x. So the first thingwe have to ask ourselves is, what doesamplitude even refer to? Well the amplitude ofa periodic function is just half the differencebetween the minimum and maximum values it takes on. So if I were to draw aperiodic function like this, and it would just go backand forth between two-- let me draw it alittle bit neater-- it goes back and forthbetween two values like that. So between thatvalue and that value. You take the difference betweenthe two, and half of that is the amplitude. Another way of thinkingabout the amplitude is how much does it swayfrom its middle position. Right over here,we have y equals negative 1/2 cosine of 3x. So what is going to bethe amplitude of this? Well, the easy wayto think about it is just what is multiplyingthe cosine function. And you could do the same thingif it was a sine function. We have negative1/2 multiplying it. So the amplitudein this situation is going to be the absolutevalue of negative 1/2, which is equal to 1/2. And you might say, well, whydo I not care about the sign? Why do I take theabsolute value of it? Well, the negative justflips the function around. It's not going tochange how much it sways between its minimumand maximum position. The other thing is,well, how is it just simply the absolutevalue of this thing? And to realize they, you just have to remember that a cosinefunction or a sine function varies between positive1 and negative 1, if it's just a simple function. So this is just multiplyingthat positive 1 or negative 1. And so if normallythe amplitude, if you didn't haveany coefficient here, if the coefficient waspositive or negative 1, the amplitude would just be 1. Now, you're changingit or you're multiplying it by this amount. So the amplitude is 1/2. Now let's thinkabout the period. So the first thingI want to ask you is, what does the periodof a cyclical function-- or periodic function,I should say-- what does the period ofa periodic function even refer to? Well let me draw some axes onthis function right over here. Let's say that this rightover here is the y-axis. That's the y-axis. And let's just say, forthe sake of argument, this is the x-axisright over here. So the period of aperiodic function is the length of thesmallest interval that contains exactly one copyof the repeating pattern of that periodic function. So what do they mean here? Well, what's repeating? So we go down and thenup just like that. Then we go downand then we go up. So in this case, the lengthof the smallest interval that contains exactly onecopy of the repeating pattern. This could be one of thesmallest repeating patterns. And so this length between hereand here would be one period. Then we could go between hereand here is another period. And there's multiple--this isn't the only pattern that you could pick. You could say, well, I'm goingto define my pattern starting here going up and thengoing down like that. So you could say that'smy smallest length. And then you wouldsee that, OK, well, if you go in thenegative direction, the next repeatingversion of that pattern is right over there. But either way you're goingto get the same length that it takes torepeat that pattern. So given that,what is the period of this functionright over here? Well, to figure out theperiod, we just take 2 pi and divide it bythe absolute value of the coefficientright over here. So we divide it by the absolutevalue of 3, which is just 3. So we get 2 pi over 3. Now we need to thinkabout why does this work? Well, if you think aboutjust a traditional cosine function, a traditional cosinefunction or a traditional sine function, it hasa period of 2 pi. If you think aboutthe unit circle, 2 pi, if you start at0, 2 pi radians later, you're back towhere you started. 2 pi radians,another 2 pi, you're back to where you started. If you go in thenegative direction, you go negative 2 pi, you'reback to where you started. For any angle here,if you go 2 pi, you're back to whereyou were before. You go negative 2 pi, you'reback to where you were before. So the periods forthese are all 2 pi. And the reason whythis makes sense is that this coefficientmakes you get to 2 pi or negative-- inthis case 2 pi-- it's going to make you getto 2 pi all that much faster. And so it gets-- your periodis going to be a lower number. It takes less length. You're going to get to 2pi three times as fast. Now you might say,well, why are you taking the absolute value here? Well, if this wasa negative number, it would get you to negative2 pi all that much faster. But either way, you're goingto be completing one cycle. So with that outof the way, let's visualize these two things. Let's actually drawnegative 1/2 cosine of 3x. So let me draw my axes here. My best attempt. So this is my y-axis. This is my x-axis. And then let me draw some--So this is 0 right over here. x is equal to 0. And let me draw x isequal to positive 1/2. I'll draw it right over here. So x is equal to positive 1/2. And we haven't shifted thisfunction up or down any. Then, if we wanted to, wecould add a constant out here, outside of the cosine function. But this is positive 1/2, or wecould just write that as 1/2. And then down here, let's saythat this is negative 1/2. And so let me draw that bound. I'm just drawingthese dotted lines so it'll becomeeasy for me to draw. And what happens when this is 0? Well cosine of 0 is 1. But we're going to multiplyit by negative 1/2. So it's going to be negative1/2 right over here. And then it's goingto start going up. It can only go inthat direction. It's bounded. It's going to start goingup, then it'll come back down and then it will get backto that original point right over here. And the question is,what is this distance? What is this length? What is this length going to be? Well, we know whatits period is. It's 2 pi over 3. It's going to get tothis point three times as fast as a traditionalcosine function. So this is goingto be 2 pi over 3. And then if you giveit another 2 pi over 3, it's going to get backto that same point again. So if you go another 2 piover 3, so in this case, you've now gone 4 pi over 3,you've completed another cycle. So that length rightover there is a period. And then you couldalso do the same thing in the negative direction. So this right over here would benegative, negative 2 pi over 3. And to visualize the amplitude,you see that it can go 1/2. Well, there's two waysto think about it. The difference between themaximum and the minimum point is 1.


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Half of that is 1/2. Or you could say that it'sgoing 1/2 in magnitude, or it's swaying 1/2 awayfrom its middle position in the positive or thenegative direction.