1. **State the problem:** We need to sketch the graph of the function $$y = \sin(4x) - 2$$ including two full periods and correct axis markings. 2. **Recall the sine function properties:** - The general form is $$y = A \sin(Bx - C) + D$$ where: - $$A$$ is the amplitude, - $$\frac{2\pi}{B}$$ is the period, - $$D$$ is the vertical shift (midline). 3. **Identify parameters for our function:** - Amplitude $$A = 1$$ (coefficient of sine) - Period $$= \frac{2\pi}{4} = \frac{\pi}{2}$$ - Midline $$y = -2$$ (vertical shift) 4. **Determine the range:** - Since amplitude is 1 and midline is -2, the function oscillates between $$-2 + 1 = -1$$ and $$-2 - 1 = -3$$. 5. **Plot key points for one period:** - Start at $$x=0$$: $$y = \sin(0) - 2 = 0 - 2 = -2$$ (midline) - At $$x = \frac{\pi}{8}$$ (quarter period): $$y = \sin(4 \cdot \frac{\pi}{8}) - 2 = \sin(\frac{\pi}{2}) - 2 = 1 - 2 = -1$$ (maximum) - At $$x = \frac{\pi}{4}$$ (half period): $$y = \sin(4 \cdot \frac{\pi}{4}) - 2 = \sin(\pi) - 2 = 0 - 2 = -2$$ (midline) - At $$x = \frac{3\pi}{8}$$ (three quarters): $$y = \sin(4 \cdot \frac{3\pi}{8}) - 2 = \sin(\frac{3\pi}{2}) - 2 = -1 - 2 = -3$$ (minimum) - At $$x = \frac{\pi}{2}$$ (full period): $$y = \sin(4 \cdot \frac{\pi}{2}) - 2 = \sin(2\pi) - 2 = 0 - 2 = -2$$ (midline) 6. **Two full periods:** - Since one period is $$\frac{\pi}{2}$$, two periods span $$x \in [0, \pi]$$. - Repeat the above pattern for $$x \in [\frac{\pi}{2}, \pi]$$. 7. **Summary:** - The graph is a sine wave with amplitude 1, period $$\frac{\pi}{2}$$, midline at $$y = -2$$. - It oscillates between -1 and -3. - Starts at (0, -2), rises to -1 at $$x=\frac{\pi}{8}$$, back to -2 at $$x=\frac{\pi}{4}$$, down to -3 at $$x=\frac{3\pi}{8}$$, and back to -2 at $$x=\frac{\pi}{2}$$. This pattern repeats for the second period. **Final answer:** The graph of $$y = \sin(4x) - 2$$ is a sine wave with amplitude 1, period $$\frac{\pi}{2}$$, midline $$y = -2$$, oscillating between -1 and -3, starting at (0, -2) and rising first.