Answer :
To write a sine function with the given properties, let's go through the details step by step:
1. Midline: The midline of the sine function is the horizontal line that the function oscillates about. For this function, the midline is [tex]\( y = 5 \)[/tex]. This means that the sine function's formula will be shifted up by 5 units.
2. Amplitude: The amplitude is the distance from the midline to the maximum or minimum value of the sine function. Here, the amplitude is 4. This means the sine function will stretch vertically by a factor of 4.
3. Period: The period of a sine function is the length along the x-axis for one complete cycle of the wave. For the standard sine function [tex]\( y = \sin(x) \)[/tex], the period is [tex]\( 2\pi \)[/tex]. We are given that the period should be [tex]\( \frac{5}{3} \)[/tex].
The standard 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,
- [tex]\( C \)[/tex] is the horizontal shift (phase shift),
- [tex]\( D \)[/tex] is the vertical shift or midline.
Given:
- Amplitude [tex]\( A = 4 \)[/tex],
- Midline [tex]\( D = 5 \)[/tex],
- Period is [tex]\( \frac{5}{3} \)[/tex].
We need to determine [tex]\( B \)[/tex] for the given period. The period of the sine function is determined by the factor [tex]\( B \)[/tex]:
[tex]\[ \text{Period} = \frac{2\pi}{B} \][/tex]
To find [tex]\( B \)[/tex], we solve for [tex]\( B \)[/tex] using the given period:
[tex]\[ \frac{2\pi}{B} = \frac{5}{3} \][/tex]
Rearranging for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{2\pi}{\frac{5}{3}} = 2\pi \cdot \frac{3}{5} = \frac{6\pi}{5} \][/tex]
Therefore, the complete sine function that fits the given properties is:
[tex]\[ y = 4 \sin \left( \frac{6\pi}{5} x \right) + 5 \][/tex]
This function has a midline of [tex]\( y = 5 \)[/tex], an amplitude of 4, and a period of [tex]\( \frac{5}{3} \)[/tex].
1. Midline: The midline of the sine function is the horizontal line that the function oscillates about. For this function, the midline is [tex]\( y = 5 \)[/tex]. This means that the sine function's formula will be shifted up by 5 units.
2. Amplitude: The amplitude is the distance from the midline to the maximum or minimum value of the sine function. Here, the amplitude is 4. This means the sine function will stretch vertically by a factor of 4.
3. Period: The period of a sine function is the length along the x-axis for one complete cycle of the wave. For the standard sine function [tex]\( y = \sin(x) \)[/tex], the period is [tex]\( 2\pi \)[/tex]. We are given that the period should be [tex]\( \frac{5}{3} \)[/tex].
The standard 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,
- [tex]\( C \)[/tex] is the horizontal shift (phase shift),
- [tex]\( D \)[/tex] is the vertical shift or midline.
Given:
- Amplitude [tex]\( A = 4 \)[/tex],
- Midline [tex]\( D = 5 \)[/tex],
- Period is [tex]\( \frac{5}{3} \)[/tex].
We need to determine [tex]\( B \)[/tex] for the given period. The period of the sine function is determined by the factor [tex]\( B \)[/tex]:
[tex]\[ \text{Period} = \frac{2\pi}{B} \][/tex]
To find [tex]\( B \)[/tex], we solve for [tex]\( B \)[/tex] using the given period:
[tex]\[ \frac{2\pi}{B} = \frac{5}{3} \][/tex]
Rearranging for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{2\pi}{\frac{5}{3}} = 2\pi \cdot \frac{3}{5} = \frac{6\pi}{5} \][/tex]
Therefore, the complete sine function that fits the given properties is:
[tex]\[ y = 4 \sin \left( \frac{6\pi}{5} x \right) + 5 \][/tex]
This function has a midline of [tex]\( y = 5 \)[/tex], an amplitude of 4, and a period of [tex]\( \frac{5}{3} \)[/tex].
