# Finding amplitude, period, phase shift on uglier functions

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## Homework Statement

State the amplitude, period, phase shift, and vertical shift:

##y=\frac{5}{2}sec\left(\frac{π}{x}-4π\right)-2##

## The Attempt at a Solution

[/B]
Amplitude:
Amplitude is equal to the absolute value of a. So the amplitude here is ##\frac{5}{2}##

Vertical shift:
Down 2.

Phase shift and Period:
This is where I'm getting thrown off and it's because of the ##\frac{π}{x}## term.
How would I go about shaking this thing to get it into a more manageable form so that I can determine the phase shift and period? From all of the problems I've seen, x has had an integer or fractional coefficient, not been in the denominator like this.

## Answers and Replies

LCKurtz
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• opus
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Holy crap that's hideous. So how would we go about determining the period/phase shift then? The given equation came up on my homework and I was at a loss.

Ray Vickson
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Dearly Missed

## Homework Statement

State the amplitude, period, phase shift, and vertical shift:

##y=\frac{5}{2}sec\left(\frac{π}{x}-4π\right)-2##

## The Attempt at a Solution

[/B]
Amplitude:
Amplitude is equal to the absolute value of a. So the amplitude here is ##\frac{5}{2}##

Vertical shift:
Down 2.

Phase shift and Period:
This is where I'm getting thrown off and it's because of the ##\frac{π}{x}## term.
How would I go about shaking this thing to get it into a more manageable form so that I can determine the phase shift and period? From all of the problems I've seen, x has had an integer or fractional coefficient, not been in the denominator like this.

I suggest you make a plot of the function, either using a graphing calculator, a spreadsheet, or some type of on-line graphing facility You will soon see why the question is meaningless: the function does not have a period and does not have a finite amplitude.

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So the homework question is useless? That would make some sense, because the problems are apparently generated with an algorithm which allows the student to do the same types of problems repeatedly.

Mark44
Mentor
So the homework question is useless?
You haven't made a typo, have you?

Is the problem supposed to be ##y=\frac{5}{2}\sec\left(\frac{x}{\pi}-4π\right)-2##? Now it's periodic, but the amplitude is still infinite.

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No sir! I checked for errors. I should’ve screenshotted it.

Gold Member
As a side question, I thought the only functions with amplitude were sinusoidal graphs? So this does have amplitude, its just infinite?

Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
As a side question, I thought the only functions with amplitude were sinusoidal graphs? So this does have amplitude, its just infinite?

No: lots of non-sinusoidal functions are periodic and have finite amplitudes. For example, try drawing a graph of the function ##f(x) = \sin x + 2 \cos 2x - 3 \sin^2 x ##. Of course, in such a case one would need to define the concept of "amplitude", but a reasonable definition might be ##\max \{f(x) \} - \min \{ f(x) \} .##

LCKurtz
Science Advisor
Homework Helper
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As a side question, I thought the only functions with amplitude were sinusoidal graphs? So this does have amplitude, its just infinite?

I think the term amplitude is usually defined for periodic functions, not just sinusoidals. But not all periodic functions have an amplitude. In physics the amplitude is the magnitude of the displacement from the neutral or zero position. In that sense, the amplitude of a secant function wouldn't make any sense, even as not in your case when it is periodic.