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

Question

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = 3 cos(2x − π)

Accepted Answer

Hello, everyone. We are asked to find the amplitude phase shift and period of the given function and sketch its graph for one period. The function we are given is Y equals seven multiplied by the cosine of in parentheses, three X minus two pi we are given a coordinate plane where the X axis is measured in pi divided by six increments. And the Y axis is measured in increments of one. Recalling that Y equals a multiplied by the cosine of BX minus C is our general format for a cosine function. I'm gonna match that to our function which is Y equals seven multiplied by the cosine of in parentheses, three X minus two pi so I'm gonna start with the amplitude. So the amplitude of our function is how high or low it goes. Amplitude can be found by finding the absolute value of the value A A in our function is seven. So the absolute value of seven is seven. So our amplitude is seven. So instead of going up to one and down to negative one, we are going to go up to positive seven and down to negative seven. Next, we'll look for the phase shift. So phase shift can be found by doing the value of C divided by the value of B. So matching this up, our C is two PI RB is three. So this means our phase shift is two pi divided by three. So instead of starting at the point 01, we will start at two pi divided by three and seven. Now that we know our amplitude and now to find the period, the period is found by doing two pi divided by B. So B in our case is a value of three. So our period is two pi divided by three. So now we can use this information to make an XY table. So to find my first X value, I know that traditionally this will start at zero. However, adding our P shift of two pi divided by three, my first X value will be at two pi divided by three and it'll have a Y value of seven because cosine functions start up and then go down to find our next X value. We need to actually first take the period which is two pi divided by three and divide that into four pieces. So two pi divided by three, divided by four is the same as multiplied by 1/4. This will give us two pie divided by which can be reduced to pi divided by six. So I'm going to be adding pi divided by six for each X value. So for my second XL U I have two pi divided by three plus pi divided by six. And once I make a common denominator, this is five pi divided by six as our second X value. And that will have a Y value of zero. Our third X value will be five pi divided by six plus pi divided by six which will be six pi divided by six, which is the same as pi. And this will have a Y value of negative seven because typically it would be negative one. But we are multiplying it by our amplitude. Next, our fourth uh sorry, our fourth X value we have pie and we're gonna add pie divided by six to it. And that would be seven pi divided by six. And that will give us a Y value of zero again. And for our last X value, seven pi divided by six plus pi divided by six is eight pi divided by six which would reduce to four pi divided by three. So four pi divided by three is the end of one period and that will end at a Y value of positive seven. So now let's get these points on my graph. So two pi divided by three as an X value has a Y value of seven, five pi divided by six has a Y value of a zero. So right on that X axis pi has a Y value of negative seven, seven pi divided by six brings us back to zero and four pi divided by three will go back up to seven for a Y value trying to connect this in a wave. We have a nice little U shape and it's in the 1st and 4th quadrants and have a nice day.

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = 3 cos(2x − π)

Key Concepts

Amplitude of a Trigonometric Function

Period of a Trigonometric Function

Phase Shift of a Trigonometric Function

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