Question

write a sine function that has an amplitude of 2, a midline of y = 3 and a period of \\(\frac{\pi}{6}\\).
answer
attempt 1 out of 2
\\(f(x) = \square\\)

Explanation:

Step1: Recall sine function form

The general form of a sine function is \( f(x) = A\sin(Bx) + k \), where \( |A| \) is the amplitude, \( \frac{2\pi}{|B|} \) is the period, and \( k \) is the midline.

Step2: Determine A (amplitude)

Given amplitude is 2, so \( A = 2 \).

Step3: Determine B (period factor)

Period is \( \frac{\pi}{6} \). Using the period formula \( \text{Period} = \frac{2\pi}{|B|} \), we solve for \( B \):

$$ \frac{\pi}{6} = \frac{2\pi}{|B|} \implies |B| = \frac{2\pi \times 6}{\pi} = 12 $$

Assume \( B = 12 \) (positive for simplicity).

Step4: Determine k (midline)

Midline is \( y = 3 \), so \( k = 3 \).

Step5: Construct the function

Substitute \( A = 2 \), \( B = 12 \), \( k = 3 \) into the general form: \( f(x) = 2\sin(12x) + 3 \).

Answer:

\( f(x) = 2\sin(12x) + 3 \)