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Let's break down the solution for the function [tex]\( y = 3.5 \sin \left(\frac{2 \pi}{365}(x-90)\right) + 12 \)[/tex]:
### Step-by-Step Analysis:
1. Understanding the Function:
- The given function is [tex]\( y = 3.5 \sin \left(\frac{2 \pi}{365}(x-90)\right) + 12 \)[/tex].
- This is a sinusoidal function, which means it represents a wave-like pattern.
2. Amplitude, Period, Phase Shift, and Vertical Shift:
- Amplitude: The coefficient of the sine function is 3.5, so the amplitude is 3.5. This means the function oscillates 3.5 units above and below the central value.
- Period: The period of the sine function is given by the coefficient of [tex]\( x \)[/tex] inside the sine function. Here, the period is [tex]\( P = \frac{2\pi}{\frac{2\pi}{365}} = 365 \)[/tex]. This means the function completes one full cycle every 365 units.
- Phase Shift: The term [tex]\( (x-90) \)[/tex] indicates a phase shift. The function is shifted to the right by 90 units.
- Vertical Shift: The +12 outside the sine function indicates that the entire function is shifted up by 12 units.
3. Range of Values:
- The sine function varies between -1 and 1.
- Therefore, [tex]\( 3.5 \sin \left(\frac{2 \pi}{365}(x-90)\right) \)[/tex] varies between -3.5 and 3.5.
- Adding 12 shifts this range to between [tex]\( 12 - 3.5 = 8.5 \)[/tex] and [tex]\( 12 + 3.5 = 15.5 \)[/tex].
4. Values for Specific X:
- To analyze the function at various points, consider [tex]\( x \)[/tex] values between 0 and 365, repeating every 365 units.
### Values of [tex]\( y \)[/tex] for a range of [tex]\( x \)[/tex] :
Here are selected values for the function [tex]\( y = 3.5 \sin \left(\frac{2 \pi}{365}(x-90)\right) + 12 \)[/tex] calculated between [tex]\( x = 0 \)[/tex] and [tex]\( x = 365 \)[/tex]:
[tex]\[ \begin{array}{c|c} x & y \\ \hline 0 & 8.50081024 \\ 90 & 12.0 \\ 182.5 & 15.50023694 \\ 273.75 & 12.0 \\ 365 & 8.50081024 \\ \end{array} \][/tex]
### Brief Conclusion:
- At [tex]\( x = 0 \)[/tex], the function starts at approximately [tex]\( y = 8.50081024 \)[/tex].
- At [tex]\( x = 90 \)[/tex], the function reaches its central value [tex]\( y = 12.0 \)[/tex].
- At [tex]\( x = 182.5 \)[/tex], the function reaches its maximum value [tex]\( y = 15.50023694 \)[/tex].
- At [tex]\( x = 273.75 \)[/tex], the function returns to the central value [tex]\( y = 12.0 \)[/tex].
- At [tex]\( x = 365 \)[/tex], the function completes one full cycle and returns to [tex]\( y = 8.50081024 \)[/tex].
This detailed analysis shows the behavior of the sinusoidal function over one period, illustrating its periodic, oscillating nature.
### Step-by-Step Analysis:
1. Understanding the Function:
- The given function is [tex]\( y = 3.5 \sin \left(\frac{2 \pi}{365}(x-90)\right) + 12 \)[/tex].
- This is a sinusoidal function, which means it represents a wave-like pattern.
2. Amplitude, Period, Phase Shift, and Vertical Shift:
- Amplitude: The coefficient of the sine function is 3.5, so the amplitude is 3.5. This means the function oscillates 3.5 units above and below the central value.
- Period: The period of the sine function is given by the coefficient of [tex]\( x \)[/tex] inside the sine function. Here, the period is [tex]\( P = \frac{2\pi}{\frac{2\pi}{365}} = 365 \)[/tex]. This means the function completes one full cycle every 365 units.
- Phase Shift: The term [tex]\( (x-90) \)[/tex] indicates a phase shift. The function is shifted to the right by 90 units.
- Vertical Shift: The +12 outside the sine function indicates that the entire function is shifted up by 12 units.
3. Range of Values:
- The sine function varies between -1 and 1.
- Therefore, [tex]\( 3.5 \sin \left(\frac{2 \pi}{365}(x-90)\right) \)[/tex] varies between -3.5 and 3.5.
- Adding 12 shifts this range to between [tex]\( 12 - 3.5 = 8.5 \)[/tex] and [tex]\( 12 + 3.5 = 15.5 \)[/tex].
4. Values for Specific X:
- To analyze the function at various points, consider [tex]\( x \)[/tex] values between 0 and 365, repeating every 365 units.
### Values of [tex]\( y \)[/tex] for a range of [tex]\( x \)[/tex] :
Here are selected values for the function [tex]\( y = 3.5 \sin \left(\frac{2 \pi}{365}(x-90)\right) + 12 \)[/tex] calculated between [tex]\( x = 0 \)[/tex] and [tex]\( x = 365 \)[/tex]:
[tex]\[ \begin{array}{c|c} x & y \\ \hline 0 & 8.50081024 \\ 90 & 12.0 \\ 182.5 & 15.50023694 \\ 273.75 & 12.0 \\ 365 & 8.50081024 \\ \end{array} \][/tex]
### Brief Conclusion:
- At [tex]\( x = 0 \)[/tex], the function starts at approximately [tex]\( y = 8.50081024 \)[/tex].
- At [tex]\( x = 90 \)[/tex], the function reaches its central value [tex]\( y = 12.0 \)[/tex].
- At [tex]\( x = 182.5 \)[/tex], the function reaches its maximum value [tex]\( y = 15.50023694 \)[/tex].
- At [tex]\( x = 273.75 \)[/tex], the function returns to the central value [tex]\( y = 12.0 \)[/tex].
- At [tex]\( x = 365 \)[/tex], the function completes one full cycle and returns to [tex]\( y = 8.50081024 \)[/tex].
This detailed analysis shows the behavior of the sinusoidal function over one period, illustrating its periodic, oscillating nature.
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