Determine an equation of the form y = a cos bx or y = a sin bx...

Question

Determine an equation of the form y = a cos bx or y = a sin bx, where b > 0, for the given graph. See Example 6.

Accepted Answer

Hey, everyone in this problem, we're asked to find the equation of the graph given below. So shown this graph looks like part of a periodic function. And it's shown in blue, we're gonna come back to the details of this graph as we work through the problem. But first, let's look at the answer choices we're given. So we have four of them. Option A Y is equal to negative two cosine of 16 X. Option by is equal to two cosine of 16 X. Option cy is equal to two sine of 16 X and option DY is equal to two cosine of six X. All right. So the first thing we're gonna look at is where our function crosses the Y axis. OK. So when X is equal to zero, in this case, the starting point of our graph, our function is at what looks like its maximum value. And because it's at its maximum value, we're gonna use the cosine equation. OK. Recall that cosine also starts at its maximum value. OK. So if we use the cosine equation, we don't have to worry about doing any sort of phase shifting of our function or anything like that. OK. So the simplest thing is to use cosine. So when we do that, our equation is going to be Y is equal to plus or minus a cosign of B multiplied by X plus C PTI. Now, this looks complicated, but we can simplify this very quickly. Now, we've just mentioned that we're using cosine. So we don't have to worry about a phase shift. Our function at X equals zero is at its maximum just like cosine. So there is no phase shift. And so our C value is going to be zero. Now, what about the devalue? OK. The devalue is a vertical shift. We look at our function, the maximum is at two, the minimum is at negative two and halfway between those is Y is equal to zero. OK. That's just the X axis. And so this graph has not been shifted up or downwards that center, the vertical center is at Y equals zero. And so D is also going to be zero. OK. So when we apply both of those, we have that our function is just going to be Y is equal to plus or minus a cosine of BX. OK. So we've already got it simplified, we need to figure out our A value and we need to figure out our B value starting with A K and A is going to be equal to our amplitude. And recall that the amplitude is the distance from the maximum to the center or from the minimum to the center. OK. It's half of the distance between the maximum and the min. So on our graph, we're gonna go to our maximum value of two and we are gonna look at the distance between that and the X axis. And alternatively, we could go from our minimum up to that X axis. That would also be the amplitude. And in this case, the maximum is two, we know that our graph is centered at Y equals zero. And so from 2 to 0, that's a distance of two. And so our amplitude is going to be equal to two. All right. So we have our amplitude along with our amplitude a value, we need to figure out the sign of our graph as well. So our regular cosine function if we think about the graph of cosine, OK, it starts at its maximum and decreases to its minimum. Before increasing. Again, our graph does the exact same. We're starting at the maximum and we're decreasing. And so the leading term is going to be positive. OK. We don't have to reflect this or flip this graph. It's just like the regular cosine graph. OK. All right. So last thing we need to do is find this B value and recall that B is related to the period of our graph. So B is going to be two pi divided by the period T. Now the period T is going to be the distance to complete one full cycle. So we look at our graph from our maximum all the way down to our minimum. We can look at the X distance between those two points. And when we do that and we can see that it's going to be six pi we start at zero, we go up to six P. We have to be a little bit careful here because this is not one full cycle. And look at our cosign curve, we're going from the maximum down to the minimum that only represents half of our cycle. And so what we've drawn on our graph in green and this line from zero all the way up to six pi is half of our period T. OK. And so what we have is that half of our period T is equal to six pi. And so our period T must be equal to 12 pi. OK. So even though we weren't shown a full cycle in our graph, we can still figure out what that period is based off of the information we were given, right? All right. Now that we know our period, we can get back to calculating BB is going to be two pi divided by 12 pi which is going to be equal to 16. So we know the amplitude A, we know B, we know the leading sine. And so our equation is going to be Y is equal to two cosine of 16 X. And that is the equation for the graph we were given, we take a look at our answer choices. We can see that this corresponds with answer choice. And B thanks everyone for watching. I hope this video helped see you in the next one.

Determine an equation of the form y = a cos bx or y = a sin bx, where b > 0, for the given graph. See Example 6.
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Key Concepts

General Form of Sine and Cosine Functions

Amplitude and Period of Trigonometric Functions

Phase Shift and Choosing Between Sine and Cosine

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