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
 
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