1. The problem asks to analyze and draw graphs of the given trigonometric functions: A: $y=6\sin(3t)$ B: $y=6\sin(6t)$ C: $y=3\sin(6t)$ D: $y=6\sin(6t+\frac{\pi}{3})$ 2. The general form of a 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). 3. The period $T$ of the sine function is given by: $$T=\frac{2\pi}{|B|}$$ 4. Let's find the amplitude and period for each function: - A: Amplitude $=6$, Period $=\frac{2\pi}{3}$ - B: Amplitude $=6$, Period $=\frac{2\pi}{6}=\frac{\pi}{3}$ - C: Amplitude $=3$, Period $=\frac{2\pi}{6}=\frac{\pi}{3}$ - D: Amplitude $=6$, Period $=\frac{\pi}{3}$, Phase shift $= -\frac{C}{B} = -\frac{\pi/3}{6} = -\frac{\pi}{18}$ (shift to the right) 5. These functions differ in amplitude, frequency (related to period), and phase shift. 6. Graphs would show sine waves with these properties: - A has lower frequency (longer period) and amplitude 6. - B and C have higher frequency (shorter period) but different amplitudes. - D is like B but shifted horizontally by $-\frac{\pi}{18}$. Final answers: A: $y=6\sin(3t)$ with amplitude 6 and period $\frac{2\pi}{3}$ B: $y=6\sin(6t)$ with amplitude 6 and period $\frac{\pi}{3}$ C: $y=3\sin(6t)$ with amplitude 3 and period $\frac{\pi}{3}$ D: $y=6\sin(6t+\frac{\pi}{3})$ with amplitude 6, period $\frac{\pi}{3}$, phase shift $-\frac{\pi}{18}$