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/2)

Accepted Answer

Find the amplitude phase shift and period of the given function. And sketches graph for one period where we have Y equals negative 12 cosine five X minus pi divided by three. Now to graph this and find all of our values. We need, we first need the standard form of cosine which is given here where we have Y equals a cosine BX minus C. In this, the opposite value of A is per amplitude. Two pi divided by B is our period. And for our face shift, we get C divided by B. Now we can first find their amplitude. We have the absolute value of negative 12 which is just 12. Our amplitude will be 12. Our period will be two pi divided by B where RB is five. Finally, our face shift will be C divided by B or pi divided by three, all divided by five, which gives us pi divided by 50. Now the phase shift tells us we will start at X equals pi divided by 15 and move to the right to graph this. We need five points. The end of our function, we'll go to pi divided by 15 plus our period, we can divide our period by four to determine where our points can be at. You have two pi divided by five all divided by four, which gives us two pi divided by 20 or pi divided by 10. We will increment our X value by pi divided by 10. Every time our first point happens at our phase shift pi divided by 15. Now our first point will always begin at the amplitude for our wife. In this case, we have a negative. So this occurs at our negative amplitude negative the 12th. Now we shift to the right by pi divided by 10. If we add pi divided by 10 to our previous X, we get pi divided by six. Now our second point will cross at Y equals zero. So we will say this is zero. Now we're crossing our line and going into the negative, which in our case, since we have a negative amplitude, we have crossed to the positive. Our next X will be four pi divided by and 12 for our Y since we're going on the positive amplitude. Now, now, since this is cosigned, we will come back down and go back to zero on our Y and we also add another pi divided by 10, we get 11 pi divided by 30 and zero. Finally, we add another pie divided by 10 to get seven pie divided by 15. And this will go back to our negative amplitude negative 12, we now have five points. We can plot onto the graph pi divided by 15 and negative pi divided by 60, which you kind of have to estimate based on this graph. It's roughly somewhere over here, four or five divided by 15, 12, 11 pi divided by 30 0. And finally seven pi divided by 15, negative 12. We can now make a curve between our points and that we now have the graph of our function with our amplitude period and face shift. OK. 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 = −3 cos (2x − π/2)

Key Concepts

Amplitude of a Trigonometric Function

Period of a Trigonometric Function

Phase Shift of a Trigonometric Function

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