Question

write a sine function that has an amplitude of 5, a midline of $y = 2$ and a period of $\frac{1}{3}$.

Explanation:

Step1: Recall the general form of a sine function

The general form of a sine function is \( y = A\sin(Bx - C)+D \), where:

• \( |A| \) is the amplitude,
• \( \frac{2\pi}{|B|} \) is the period,
• \( \frac{C}{B} \) is the phase shift (we can assume \( C = 0 \) for no phase shift),
• \( D \) is the vertical shift (which determines the midline).

Step2: Determine the value of \( A \)

Given the amplitude is 5, so \( A = 5 \) (we can take the positive value for a standard sine function without reflection for now).

Step3: Determine the value of \( B \)

The period is given as \( \frac{1}{3} \). We know that the period \( P=\frac{2\pi}{|B|} \). So we set up the equation:
\( \frac{1}{3}=\frac{2\pi}{|B|} \)
To solve for \( |B| \), we can cross - multiply: \( |B|=2\pi\times3 = 6\pi \). We can take \( B = 6\pi \) (since we can have no reflection, \( B>0 \)).

Step4: Determine the value of \( D \)

The midline is \( y = 2 \). The midline of the sine function \( y = A\sin(Bx - C)+D \) is \( y = D \). So \( D = 2 \).

Step5: Write the sine function

Since we assumed \( C = 0 \) (no phase shift), substituting \( A = 5 \), \( B = 6\pi \), \( C = 0 \) and \( D = 2 \) into the general form \( y=A\sin(Bx - C)+D \), we get:
\( y = 5\sin(6\pi x)+2 \)

Answer:

\( y = 5\sin(6\pi x)+2 \)