- #1

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I know the answer is pie; -3pie/8, but I don't understand the process could someone explain how the period and phase shift are found in these functions.

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- Thread starter Jess048
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- #1

- 9

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I know the answer is pie; -3pie/8, but I don't understand the process could someone explain how the period and phase shift are found in these functions.

- #2

Integral

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Could you share with us the general forumla given in your text or notes for this type of function?

- #3

HallsofIvy

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Do you know the definitions of "phase shift" and period? What is the period of tan(x)?

- #4

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The period of function y = tan k0 is pie/k where k>o. I'm not sure about the phase shift.

- #5

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Consider y=x^2

y=(x-2.5)^2 is the same function, shifted to the right 2.5 units.

y=(2x-6)^2 would have to first be written as

y=(2(x-3))^2

This is the function y=x^2 shifted 3 units to the right. The 2 does something else to the function (stretches it vertically in this case.)

Can you get (x-#) in your problem?

- #6

HallsofIvy

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One period starts at x= 0 and ends at [itex]x= \pi[/itex].

Okay, one period of [itex]-4tan((1/2)x- 3\pi/8)[/itex] "starts" when [itex](1/2)x- 3\pi/8= 0[/itex] and ends when [itex](1/2)x- 3\pi/8= \pi[/itex]. What is the "starting" value of x (the phase shift) and what is the difference between the two values of x (the period)?

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