MHB -7.8.1 Amp, Period, PS, VS of 3cos(\pi x-2)+5

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The discussion focuses on finding the amplitude, period, phase shift (PS), and vertical shift (VS) of the function y=3cos(πx-2)+5. The amplitude is determined to be 3, the vertical shift is 5, and the period is calculated as 2 using the formula T=2π/ω. The phase shift is found to be 2/π. Participants confirm the correctness of the equations and calculations throughout the discussion, ensuring clarity on the parameters involved.
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Find amplitude, period, PS, VS. graph 2 periods of
$y=3\cos(\pi x-2)+5$

ok I think these are the plug ins we use
$Y=A\cos\left[\omega\left(x-\dfrac{x \phi}{\omega} \right)\right]+B $
or
$A\cos\left(\omega x-\phi\right)+B$
A=amplitude B=VS or veritical shift
$T = \dfrac{2\pi}{\omega-\phi}$
$PS = 0$ assumed here

ok just want to see if I have these plug in eq right, different books use different symbols
 
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karush said:
Find amplitude, period, PS, VS. graph 2 periods of
$y=3\cos(\pi x-2)+5$

ok I think these are the plug ins we use
$Y=A\cos\left[\omega\left(x-\dfrac{x \phi}{\omega} \right)\right]+B $
or
$A\cos\left(\omega x-\phi\right)+B$
A=amplitude B=VS or veritical shift
$T = \dfrac{2\pi}{\omega-\phi}$
$PS = 0$ assumed here

ok just want to see if I have these plug in eq right, different books use different symbols
Use [math]Y = A\cos\left(\omega x-\phi\right)+B[/math] or [math]Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B[/math]. (You had one too many x's in your first equation.)

-Dan
 
$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$
then for $y=3\cos(\pi x-2)+5$
$A=3 \quad \omega=\pi \quad \phi=2 \quad B=5$
before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega} $ and $PS=\dfrac{\phi}{\omega}$

 
karush said:
$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$
then for $y=3\cos(\pi x-2)+5$
$A=3 \quad \omega=\pi \quad \phi=2 \quad B=5$
before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega} $ and $PS=\dfrac{\phi}{\omega}$
Yup. :)

-Dan
 

before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega} $ and $PS=\dfrac{\phi}{\omega}$
so then
$T=\dfrac{2\pi}{\pi}=2$ and $PS=\dfrac{2}{\pi}$
kinda ? on PS
So T is Period?

 
karush said:

before the plug...:unsure:
where $T=\dfrac{2\pi}{\omega} $ and $PS=\dfrac{\phi}{\omega}$
so then
$T=\dfrac{2\pi}{\pi}=2$ and $PS=\dfrac{2}{\pi}$
kinda ? on PS
So T is Period?
Yes. You have it right.

-Dan
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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