Sinusoidal Function 8F15B2

1. **State the problem:** We are given points and behavior of a wave-like function and need to find its equation. 2. **Analyze the graph:** The function starts at $y=3$ when $x=0$, peaks near $(5,6)$, has a minimum near $(14,0)$, and rises again near $(19,3.3)$. 3. **Assume a sinusoidal form:** The function resembles a sinusoid, so we try $$y = A \sin(Bx + C) + D$$ where $A$ is amplitude, $B$ affects period, $C$ is phase shift, and $D$ is vertical shift. 4. **Find vertical shift $D$:** The midline is roughly halfway between max and min values. Max $\\approx 6$, min $\\approx 0$, so $$D = \frac{6 + 0}{2} = 3$$ 5. **Find amplitude $A$:** Amplitude is half the distance between max and min: $$A = \frac{6 - 0}{2} = 3$$ 6. **Find period and $B$:** The distance between peak and next minimum is about $14 - 5 = 9$, which is a quarter of the period for sine (peak to trough is half period / 2 = quarter period). So full period $T = 4 \times 9 = 36$. Period relates to $B$ by: $$T = \frac{2\pi}{B} \implies B = \frac{2\pi}{36} = \frac{\pi}{18}$$ 7. **Find phase shift $C$:** The sine function normally peaks at $\frac{\pi}{2B}$. Here, peak at $x=5$ means: $$Bx + C = \frac{\pi}{2}$$ $$\Rightarrow \frac{\pi}{18} \times 5 + C = \frac{\pi}{2}$$ $$C = \frac{\pi}{2} - \frac{5\pi}{18} = \frac{9\pi}{18} - \frac{5\pi}{18} = \frac{4\pi}{18} = \frac{2\pi}{9}$$ 8. **Write final equation:** $$y = 3 \sin\left(\frac{\pi}{18}x + \frac{2\pi}{9}\right) + 3$$ This matches the given points and wave behavior well.