1. **State the problem:** We are given a sine function of the form $f(x) = A \sin(Bx - C)$ and a graph with specific points: amplitude 1, zero at $x = -\frac{\pi}{6}$, peak at $x = \frac{\pi}{2}$ with $y=1$, zero at $x = \frac{7\pi}{6}$, minimum at $x = \frac{11\pi}{6}$ with $y=-1$, and zero again at $x = \frac{5\pi}{2}$. 2. **Identify amplitude $A$:** Amplitude is the maximum absolute value of the sine wave. Given the peak is 1 and minimum is -1, amplitude $A = 1$. 3. **Determine the period and $B$:** The period $T$ of $f(x) = A \sin(Bx - C)$ is given by: $$T = \frac{2\pi}{B}$$ From the graph, one full cycle starts at the zero crossing $x = -\frac{\pi}{6}$ and ends at the next zero crossing after one full period at $x = \frac{5\pi}{2}$. Calculate the period: $$T = \frac{5\pi}{2} - \left(-\frac{\pi}{6}\right) = \frac{5\pi}{2} + \frac{\pi}{6} = \frac{15\pi}{6} + \frac{\pi}{6} = \frac{16\pi}{6} = \frac{8\pi}{3}$$ Use the period formula to find $B$: $$B = \frac{2\pi}{T} = \frac{2\pi}{\frac{8\pi}{3}} = \frac{2\pi \times 3}{8\pi} = \frac{6}{8} = \frac{3}{4}$$ 4. **Determine phase shift $C$:** The sine function normally starts at zero at $x=0$ with no phase shift. Here, the sine wave crosses zero at $x = -\frac{\pi}{6}$, so the phase shift moves the graph right by $\frac{\pi}{6}$. The phase shift $\phi$ is given by: $$\phi = \frac{C}{B}$$ Since the zero crossing is shifted to $x = -\frac{\pi}{6}$, the inside of sine is zero at this point: $$Bx - C = 0 \implies C = Bx = \frac{3}{4} \times \left(-\frac{\pi}{6}\right) = -\frac{3\pi}{24} = -\frac{\pi}{8}$$ 5. **Write the function:** $$f(x) = \sin\left(\frac{3}{4}x - \left(-\frac{\pi}{8}\right)\right) = \sin\left(\frac{3}{4}x + \frac{\pi}{8}\right)$$ 6. **Verify with given points:** At $x = \frac{\pi}{2}$ (peak), inside sine is: $$\frac{3}{4} \times \frac{\pi}{2} + \frac{\pi}{8} = \frac{3\pi}{8} + \frac{\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2}$$ Since $\sin(\frac{\pi}{2}) = 1$, this matches the peak. **Final answer:** $$f(x) = \sin\left(\frac{3}{4}x + \frac{\pi}{8}\right)$$