Answer :
To write a sine function with the given characteristics, let's break down the necessary components step-by-step.
1. Midline:
The midline of the sine function is the vertical shift. In this case, it is [tex]\( y = 2 \)[/tex]. So, the vertical shift [tex]\( D \)[/tex] is 2.
2. Amplitude:
The amplitude is the maximum distance from the midline to the peak or trough of the sine wave. Here, the amplitude [tex]\( A \)[/tex] is 5.
3. Period:
The period of the sine function is the length of one complete cycle of the sine wave. The given period is [tex]\( \frac{7}{4} \)[/tex].
The general form of a sine function is:
[tex]\[ y = A \sin(B(x - C)) + D \][/tex]
Where:
- [tex]\( A \)[/tex] is the amplitude
- [tex]\( B \)[/tex] affects the period with the relation [tex]\( \text{Period} = \frac{2\pi}{B} \)[/tex]
- [tex]\( C \)[/tex] is the horizontal shift (not specified in this problem, so we assume [tex]\( C = 0 \)[/tex])
- [tex]\( D \)[/tex] is the vertical shift (midline)
4. Determining [tex]\( B \)[/tex]:
To determine [tex]\( B \)[/tex], we use the relationship between [tex]\( B \)[/tex] and the period:
[tex]\[ \text{Period} = \frac{2\pi}{B} \][/tex]
Given the period is [tex]\( \frac{7}{4} \)[/tex]:
[tex]\[ \frac{7}{4} = \frac{2\pi}{B} \][/tex]
Solving for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{2\pi}{\frac{7}{4}} = \frac{2\pi \times 4}{7} = \frac{8\pi}{7} \][/tex]
Thus, [tex]\( B = \frac{8\pi}{7} \)[/tex].
Combining all the components, the sine function can be written as:
[tex]\[ y = 5 \sin\left( \frac{8\pi}{7} x \right) + 2 \][/tex]
So, the final sine function that has a midline of [tex]\( y = 2 \)[/tex], an amplitude of 5, and a period of [tex]\( \frac{7}{4} \)[/tex] is:
[tex]\[ y = 5 \sin\left( \frac{8\pi}{7} x \right) + 2 \][/tex]
Additionally, the calculated value for [tex]\( B \)[/tex] is:
[tex]\[ B = \frac{8\pi}{7} \approx 3.5903916041026207 \][/tex]
Therefore, the sine function is:
[tex]\[ y = 5 \sin(3.5903916041026207 \cdot x) + 2 \][/tex]
1. Midline:
The midline of the sine function is the vertical shift. In this case, it is [tex]\( y = 2 \)[/tex]. So, the vertical shift [tex]\( D \)[/tex] is 2.
2. Amplitude:
The amplitude is the maximum distance from the midline to the peak or trough of the sine wave. Here, the amplitude [tex]\( A \)[/tex] is 5.
3. Period:
The period of the sine function is the length of one complete cycle of the sine wave. The given period is [tex]\( \frac{7}{4} \)[/tex].
The general form of a sine function is:
[tex]\[ y = A \sin(B(x - C)) + D \][/tex]
Where:
- [tex]\( A \)[/tex] is the amplitude
- [tex]\( B \)[/tex] affects the period with the relation [tex]\( \text{Period} = \frac{2\pi}{B} \)[/tex]
- [tex]\( C \)[/tex] is the horizontal shift (not specified in this problem, so we assume [tex]\( C = 0 \)[/tex])
- [tex]\( D \)[/tex] is the vertical shift (midline)
4. Determining [tex]\( B \)[/tex]:
To determine [tex]\( B \)[/tex], we use the relationship between [tex]\( B \)[/tex] and the period:
[tex]\[ \text{Period} = \frac{2\pi}{B} \][/tex]
Given the period is [tex]\( \frac{7}{4} \)[/tex]:
[tex]\[ \frac{7}{4} = \frac{2\pi}{B} \][/tex]
Solving for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{2\pi}{\frac{7}{4}} = \frac{2\pi \times 4}{7} = \frac{8\pi}{7} \][/tex]
Thus, [tex]\( B = \frac{8\pi}{7} \)[/tex].
Combining all the components, the sine function can be written as:
[tex]\[ y = 5 \sin\left( \frac{8\pi}{7} x \right) + 2 \][/tex]
So, the final sine function that has a midline of [tex]\( y = 2 \)[/tex], an amplitude of 5, and a period of [tex]\( \frac{7}{4} \)[/tex] is:
[tex]\[ y = 5 \sin\left( \frac{8\pi}{7} x \right) + 2 \][/tex]
Additionally, the calculated value for [tex]\( B \)[/tex] is:
[tex]\[ B = \frac{8\pi}{7} \approx 3.5903916041026207 \][/tex]
Therefore, the sine function is:
[tex]\[ y = 5 \sin(3.5903916041026207 \cdot x) + 2 \][/tex]
