Sure! Let's break down how to construct a sine function with the given properties step by step.
### Step 1: Understanding the general form of a sine function
The general form of a sine function is:
[tex]\[ y = A \sin(Bx - C) + D \][/tex]
where:
- [tex]\(A\)[/tex] is the amplitude.
- [tex]\(B\)[/tex] relates to the period.
- [tex]\(C\)[/tex] is the horizontal shift (phase shift).
- [tex]\(D\)[/tex] is the vertical shift (midline).
### Step 2: Determine the amplitude [tex]\(A\)[/tex]
The amplitude [tex]\(A\)[/tex] is given as 2. This means the sine wave will have a maximum value of 2 units above the midline and a minimum value of 2 units below the midline.
[tex]\[ A = 2 \][/tex]
### Step 3: Determine the midline [tex]\(D\)[/tex]
The midline [tex]\(y = 5\)[/tex] tells us the vertical shift [tex]\(D\)[/tex] is 5. This is the value around which the sine wave oscillates.
[tex]\[ D = 5 \][/tex]
### Step 4: Calculate [tex]\(B\)[/tex] using the given period
The period (the length of one complete cycle of the sine wave) is given as [tex]\(\frac{2}{3}\)[/tex]. The relationship between the period and [tex]\(B\)[/tex] is:
[tex]\[ \text{Period} = \frac{2\pi}{B} \][/tex]
Given the period [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[ \frac{2}{3} = \frac{2\pi}{B} \][/tex]
Solving for [tex]\(B\)[/tex]:
[tex]\[ B = \frac{2\pi}{\frac{2}{3}} = 2\pi \times \frac{3}{2} = 3\pi \][/tex]
### Step 5: Determine the horizontal shift [tex]\(C\)[/tex]
There is no horizontal shift mentioned, so we set [tex]\(C = 0\)[/tex].
[tex]\[ C = 0 \][/tex]
### Step 6: Construct the sine function
Now, putting it all together, the sine function with the given properties is:
[tex]\[ y = 2 \sin(3\pi x) + 5 \][/tex]
This function has:
- An amplitude of 2.
- A midline of [tex]\(y = 5\)[/tex].
- A period of [tex]\(\frac{2}{3}\)[/tex].
Thus, the required sine function is:
[tex]\[ y = 2 \sin(3\pi x) + 5 \][/tex]
