1. **State the problem:** We need to write a sine function with the following characteristics: - Midline (vertical shift) at $y=3$ - Amplitude of 55 - Period of $\frac{7\pi}{6}$ - Horizontal shift (phase shift) of $\frac{7\pi}{12}$ to the left 2. **Recall the general sine function formula:** $$y = A \sin(B(x - C)) + D$$ where: - $A$ is the amplitude - $B$ affects the period - $C$ is the horizontal shift - $D$ is the vertical shift (midline) 3. **Calculate $B$ from the period:** The period $P$ is related to $B$ by: $$P = \frac{2\pi}{B}$$ Given $P = \frac{7\pi}{6}$, solve for $B$: $$B = \frac{2\pi}{P} = \frac{2\pi}{\frac{7\pi}{6}} = 2\pi \times \frac{6}{7\pi} = \frac{12}{7}$$ 4. **Determine the horizontal shift $C$:** The shift is $\frac{7\pi}{12}$ to the left, so: $$C = -\frac{7\pi}{12}$$ 5. **Write the function:** Substitute $A=55$, $B=\frac{12}{7}$, $C=-\frac{7\pi}{12}$, and $D=3$: $$y = 55 \sin\left(\frac{12}{7}\left(x - \left(-\frac{7\pi}{12}\right)\right)\right) + 3 = 55 \sin\left(\frac{12}{7}\left(x + \frac{7\pi}{12}\right)\right) + 3$$ **Final answer:** $$y = 55 \sin\left(\frac{12}{7}\left(x + \frac{7\pi}{12}\right)\right) + 3$$