In Exercises 43–52, determine the amplitude, period, and phase sh...

Question

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function.y = 2 cos (2πx + 8π)

Accepted Answer

Find the amplitude phase shift and period of the given function and sketch its graph for one period where we have Y equals 16 cosine of four pi X plus two pi. Now to graph this, we first need the standard form of a cosine equation. This is given by Y equals a cosine of BX minus C where A is our amplitude B is a number that multiplies our X and C is going to be a shift. Now, our period is given by two pi divided by B our face shift. It's also given by C divided by B I know our equation is Y equals 16 cosine four pi X plus two pi that makes our amplitude, the number of multiplying our cosine at 16, our period then will be two pi divided by B. We are B. This time is for a pint, meaning we have two pi divided by four pi which simplifies to one half. Finally, we have our phase shift which is C divided by B RC is two PI and this is negative divided by four pi which gives us negative one half. So we're given our amplitude period and patient no we can graph our function. The face shift tells us that we will start our period at negative one half. So we can make five key points for this graph. We're starting at negative one half and at negative one half, our first point will be our amplitude. Our first output will be our amplitude, I should say 16. Now, since our period is one half, we want to divide this period by fourth, this will help us find our five key points that is one half divided by four, which is 18. So we will actually move our point by one half every time to the right. So our next point will occur a negative one half plus 1/8 or negative three eights. Now the second point will be Y equals zero because of how the cosine graph works. Our next point will be shifted to the right by 1/ which occurs at negative 1/4. That will be our negative amplitude negative 16. Our next point shift to the right by 1/8. Again, we get negative 180. And finally our fifth point, we will add another 1/8 we get zero and our positive Amplex begin 16. Now we can plot all of our points you have and they, the one half and N 380. Can we have negative 1/4 negative 16, negative 180. And finally zero and 16, we can then make a cosine curve between these five points? Ok. I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function.y = 2 cos (2πx + 8π)

Key Concepts

Amplitude

Period

Phase Shift

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