1. **State the problem:** Graph the function $$y = -3 + 2 \sin x$$ over a two-period interval. 2. **Recall the formula and properties:** - The general sine function is $$y = A \sin(Bx + C) + D$$ where: - $$A$$ is the amplitude (height of peaks), - $$B$$ affects the period $$T = \frac{2\pi}{|B|}$$, - $$C$$ is the phase shift, - $$D$$ is the vertical shift. 3. **Identify parameters:** - Amplitude $$A = 2$$ - Frequency multiplier $$B = 1$$ so period $$T = \frac{2\pi}{1} = 2\pi$$ - Vertical shift $$D = -3$$ - No phase shift $$C = 0$$ 4. **Determine interval:** - Two periods means interval length $$2 \times 2\pi = 4\pi$$ - Choose interval $$[0, 4\pi]$$ for graphing. 5. **Calculate key points:** - Sine function key points in one period $$[0, 2\pi]$$ are at $$x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$$ with sine values $$0, 1, 0, -1, 0$$ respectively. - For two periods $$[0, 4\pi]$$, key points are: - $$x=0, y = -3 + 2 \times 0 = -3$$ - $$x=\frac{\pi}{2}, y = -3 + 2 \times 1 = -1$$ - $$x=\pi, y = -3 + 2 \times 0 = -3$$ - $$x=\frac{3\pi}{2}, y = -3 + 2 \times (-1) = -5$$ - $$x=2\pi, y = -3 + 2 \times 0 = -3$$ - $$x=\frac{5\pi}{2}, y = -3 + 2 \times 1 = -1$$ - $$x=3\pi, y = -3 + 2 \times 0 = -3$$ - $$x=\frac{7\pi}{2}, y = -3 + 2 \times (-1) = -5$$ - $$x=4\pi, y = -3 + 2 \times 0 = -3$$ 6. **Plot points and graph:** - The graph oscillates between $$-5$$ and $$-1$$ with midline at $$y = -3$$. - The sine wave completes two full cycles over $$[0, 4\pi]$$. Final answer: The function $$y = -3 + 2 \sin x$$ has amplitude 2, period $$2\pi$$, vertical shift -3, and is graphed over $$[0, 4\pi]$$ with key points as calculated above.