1. **Problem Statement:** Identify the graph characteristics of the given sine functions: A: $y=8\sin\left(2t+\frac{\pi}{4}\right)$ B: $y=4\sin(2t)$ C: $y=8\sin(t)$ D: $y=8\sin(2t)$ 2. **Formula and Key Concepts:** The general sine function is $y=A\sin(Bt+C)$ where: - $A$ is the amplitude (height of peaks), - $B$ affects the period (length of one cycle), - $C$ is the phase shift (horizontal shift). The period $T$ is given by $T=\frac{2\pi}{|B|}$. 3. **Analyze each function:** - A: Amplitude $A=8$, frequency factor $B=2$, phase shift $C=\frac{\pi}{4}$. - Period $T=\frac{2\pi}{2}=\pi$. - Phase shift $= -\frac{C}{B} = -\frac{\pi/4}{2} = -\frac{\pi}{8}$ (shift left). - B: Amplitude $A=4$, $B=2$, no phase shift. - Period $T=\pi$. - C: Amplitude $A=8$, $B=1$, no phase shift. - Period $T=2\pi$. - D: Amplitude $A=8$, $B=2$, no phase shift. - Period $T=\pi$. 4. **Summary of graphs:** - Function A: High amplitude 8, period $\pi$, shifted left by $\frac{\pi}{8}$. - Function B: Lower amplitude 4, period $\pi$, no shift. - Function C: High amplitude 8, longer period $2\pi$, no shift. - Function D: High amplitude 8, period $\pi$, no shift. 5. **Matching functions to graphs:** - Function 1: Could be $y=8\sin(2t+\frac{\pi}{4})$ (A) due to phase shift and amplitude. - Function 2: $y=4\sin(2t)$ (B) due to smaller amplitude. - Function 3: $y=8\sin(t)$ (C) due to longer period. - Function 4: $y=8\sin(2t)$ (D) due to amplitude 8 and period $\pi$ without shift. Final answer: Function 1 = A Function 2 = B Function 3 = C Function 4 = D