1. **Problem (a):** Find $A$, $B$, and $n$ for the curve $y = A + B \sin(nt)$ given the sine wave starts at $y=6$ when $t=0$ and oscillates with zero crossings near $\frac{\pi}{4}$ and $\frac{3\pi}{4}$. 2. **Step 1:** Identify the vertical shift $A$. Since the wave oscillates around $y=6$, the midline is $A=6$. 3. **Step 2:** Find amplitude $B$. The wave oscillates between $y=9$ (max) and $y=3$ (min), so amplitude $B = \frac{9-3}{2} = 3$. 4. **Step 3:** Determine frequency $n$. Zero crossings occur at $t=\frac{\pi}{4}$ and $t=\frac{3\pi}{4}$, so half a period is $\frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}$. Thus, full period $T = 2 \times \frac{\pi}{2} = \pi$. Frequency $n = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2$. 5. **Answer (a):** $A=6$, $B=3$, $n=2$. Equation: $$y = 6 + 3 \sin(2t)$$ 6. **Problem (b):** Find the range and equation of the curve oscillating between approximately $-4$ and $1$ with zero crossing near $\frac{\pi}{3}$. 7. **Step 1:** Midline $A = \frac{1 + (-4)}{2} = \frac{-3}{2} = -1.5$. 8. **Step 2:** Amplitude $B = \frac{1 - (-4)}{2} = \frac{5}{2} = 2.5$. 9. **Step 3:** Zero crossing at $t=\frac{\pi}{3}$ means the sine argument is zero there: $nt - \phi = 0$ at $t=\frac{\pi}{3}$. Assume no phase shift $\phi=0$, then $n \times \frac{\pi}{3} = 0$ is impossible, so phase shift exists. 10. **Step 4:** Since zero crossing is at $t=\frac{\pi}{3}$, phase shift $\phi = n \times \frac{\pi}{3}$. Assume $n=1$ for simplicity, then $\phi = \frac{\pi}{3}$. Equation: $$y = -1.5 + 2.5 \sin\left(t - \frac{\pi}{3}\right)$$ 11. **Answer (b):** Range $= [-4,1]$, equation: $$y = -1.5 + 2.5 \sin\left(t - \frac{\pi}{3}\right)$$ 12. **Problem (c):** Find the range and equation of the curve oscillating between approximately $-5$ and $4$ over intervals marked $2\pi,4\pi,6\pi,8\pi,10\pi,12\pi$. 13. **Step 1:** Midline $A = \frac{4 + (-5)}{2} = \frac{-1}{2} = -0.5$. 14. **Step 2:** Amplitude $B = \frac{4 - (-5)}{2} = \frac{9}{2} = 4.5$. 15. **Step 3:** Period $T$ is from $0$ to $12\pi$ for multiple cycles. Count cycles: from $0$ to $12\pi$ is 6 cycles if period $T=2\pi$. So frequency $n = \frac{2\pi}{T} = 1$. 16. **Step 4:** Assume no phase shift, equation: $$y = -0.5 + 4.5 \sin(t)$$ 17. **Answer (c):** Range $= [-5,4]$, equation: $$y = -0.5 + 4.5 \sin(t)$$