In Exercises 17–30, determine the amplitude, period, and phase sh...

Question

In Exercises 17–30, determine the amplitude, period, and phase shift of each function. Then graph one period of the function.y = −2 sin(2πx + 4π)

Accepted Answer

Welcome back. I am so glad you're here. We're asked to identify the amplitude, the phase shift and the period of the given sine trigonometric function then sketch its graph by considering only one period. Our given function is Y equals negative five sign of the quantity of two pi X plus six pi. Then we're given a graph on which we can draw our function. We have a vertical Y axis, a horizontal AX axis, they come together at the origin in the middle and then in the background is a faint grid showing each unit along the X and Y axes. All right, looking at our function, we see that this is in the format of Y equals a sign of the quantity of B X minus C. And we can identify our A B and C terms. Here A is the one in front of sign being multiplied by it. So A here is negative five B is the term being multiplied by the X. So here that's two pi and C a little bit different C is being subtracted from B X. And here we have a plus six pi. So that means our C term is going to be the opposite sign. It's going to be a negative six pi and now that we have our A B and C terms, we can relatively quickly figure out our amplitude phase shift and period. So for our amplitude, we recall from previous lessons, that is the absolute value of A here, our A is negative five. So it's the absolute value of negative five which is five. Now, for our phase shift, we recall from previous lessons that the phase shift is C divided by B C is negative six pi divided by B is two pi and negative six pi divided by two pi is negative three. Now four hour period, the period is given to us by two pi divided by B and that's two pi divided by our B is two pi. So two pi divided by two pi is one, our period is one. All right, we did those first three things asked for. Next, we are asked to sketch the graph. Now we recalling previous lessons and the steps given in the book for sketching the graph of a sign function. The first step was to find the amplitude phase shift in period. Who we got that one done. Now, step two is going to be to find our five key X values that's going to be our X intercepts and our maximum and minimum. The first X value that we're looking for is at the starting point. So we'll call this our X sub one and the starting point is going to be where X equals the phase shift. So it's going to be at negative three. Now to find all of the subsequent X values, we are going to add our period divided by four. And here the period is one. So one divided by four, we're going to add 1/4 to our starting point and then each subsequent X value to get the remaining four key X values. So for X sub two, we're taking our X sub one which is negative three and adding the period divided by four, which is 1/4 and negative three plus 1/4 is negative 11 4th. For X sub three, we take the negative 11 4th add 1/4 and that gets us simplified a negative five halves. For X of four, we take our negative five halves, add 1/4 and that gets us negative 9/4. And for our final X sub five, we take negative 9/4 add 1/4 and we get negative two. So there are key X values. Step three is to find the corresponding Y values that go with those X values. So there are a couple ways to do this. One is to just plug each X value in to our given function. So for the Y sub one, you would plug negative three in for X in the function, plug that into the calculator and you would get a Y value of zero. But we already know this because our X sub one, X sub three and X sub five, those are all X intercepts. So the Y values are going to be zero for each of those for the Y sub two and the Y sub 41 will be our maximum and one will be our minimum. So let's take a look at those. Let's start with Y sub two for Y sub two. If you plugged the X sub two value in for X in the function, you would get an answer of negative five. But we can know this without plugging that into the calculator. So for our maximum and minimum, we're going to take our amplitude and that's going to be the distance from the X axis. So our amplitude is five, but we have a negative five here, meaning that this is our minimum. But we know that by looking at our function and we see that our, a value is negative, that means this is reflected over the X axis and our minimum will be first rather than our maximum. OK? Now our Y sub three, as we said that XX sub three is an X intercept. So our Y sub three is going to be zero. You can always check it plugging in that X sub three value. But we don't need to our X, our, our Y sub four is going to be at our maximum. So that's the amplitude above the X axis. So that's a positive five and Y sub five again, that's an X intercept. So that's going to be zero. All right. Now we just need step four is to graph. So graphing each of these five key points from the origin for the first point negative 30, we go to the left three units and market for the second point negative 11 4th negative five from the origin, we go to the left 11 4th of a unit. And then we go down five units for the third point negative five halves, zero. We go from the origin to the left five halves of a unit and mark it for the fourth point negative 9/4 5 from the origin. We go to the left five or 9/4 of the unit. And then we go up five for the fifth point from the origin, we go to the left two units and then nowhere for the Y value because that point is negative to zero. Now we connect the dots as smoothly as we can connect the points. And there we have our sign function. Well done. We'll catch you on the next one.

In Exercises 17–30, determine the amplitude, period, and phase shift of each function. Then graph one period of the function.y = −2 sin(2πx + 4π)

Key Concepts

Amplitude

Period

Phase Shift

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