MHB 7.t.27 Write an equation for a sinusoidal graph with the following properties:

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The discussion focuses on deriving the equation for a sinusoidal graph with specified properties: an amplitude of -3, a period of \(\frac{2\pi}{3}\), and a phase shift of -\(\frac{\pi}{4}\). The correct equation is derived as \(y = -3\sin\left(3\left(x + \frac{\pi}{4}\right)\right)\), emphasizing that the amplitude is expressed as a positive value regardless of the sign in front. Clarification is provided that the amplitude should be represented as the absolute value, which is 3, and the phase shift's sign is reversed in the equation. The final equation accurately reflects the properties given, resolving initial confusion about the amplitude's negative value.
karush
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$\tiny{7.t.27}$
$\textsf{Write an equation for a sinusoidal graph with the following properties:}\\$
$$A=-3,
\textsf{Period}=\frac{2\pi}{3},
\textsf{Phase Shift}=-\frac{\pi}{4}$$
\begin{align*}\displaystyle
A&=-3\\
T&=\frac{2\pi}{3}=\frac{2\pi}{\omega}\\
\omega&=3\\
PS&=-\frac{\pi}{4}=\frac{\phi}{\omega}\\
\phi&=\pi
\end{align*}
\begin{align*}\displaystyle
y_{27}&=A\sin{\left[\omega\left(x-\frac{\phi}{\omega} \right)\right]}
\end{align*}
$\textit{so then}$
\begin{align*}\displaystyle
Y_{27}&=-3\sin{\left[3\left(x-\frac{\pi}{12} \right)\right]}
\end{align*}

hopefully:confused:
 
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The amplitude of a sinusoidal function is one-half the vertical distance between the minimum and maximum, and so should never given as a negative value. Given:

$$y=A\sin\left(\omega(x-\phi)\right)\tag{1}$$

The amplitude is then $|A|$. Thus, both of the following have an amplitude of 3 units:

$$y=3\sin(x)$$

$$y=-3\sin(x)$$

You have the correct angular velocity $\omega$, but in (1) the phase shift is $\phi$ and so $\phi$ will have the opposite sign of the value given for the phase shift:

$$y=3\sin\left(3\left(x-\left(-\frac{\pi}{4}\right)\right)\right)=3\sin\left(3\left(x+\frac{\pi}{4}\right)\right)$$
 
View attachment 6987

here is what was given...
 

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karush said:
here is what was given...

My mind must've been on something other than this problem when I posted last night...$A=-3$ is not saying the amplitude is -3...so what I would give is:

$$y=-3\sin\left(3\left(x+\frac{\pi}{4}\right)\right)$$

Sorry for the confusion. :D
 
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