1. **State the problem:** Graph the function $$y = 3 \cos(4x + \pi)$$ over one period and find key coordinates. 2. **Identify the formula and period:** The general cosine function is $$y = A \cos(Bx + C)$$ where: - Amplitude $$A = 3$$ - Angular frequency $$B = 4$$ - Phase shift $$-\frac{C}{B} = -\frac{\pi}{4}$$ The period $$T = \frac{2\pi}{|B|} = \frac{2\pi}{4} = \frac{\pi}{2}$$. 3. **Determine the interval for one period:** Since the period is $$\frac{\pi}{2}$$, the interval for one period starting at the phase shift is: $$x \in \left[-\frac{\pi}{4}, -\frac{\pi}{4} + \frac{\pi}{2}\right] = \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$$. 4. **Calculate key points:** We evaluate $$y$$ at four key points within one period: start, quarter, half, and three-quarters of the period. - At $$x = -\frac{\pi}{4}$$ (start): $$y = 3 \cos\left(4\left(-\frac{\pi}{4}\right) + \pi\right) = 3 \cos(-\pi + \pi) = 3 \cos(0) = 3$$ - At $$x = -\frac{\pi}{4} + \frac{\pi}{8} = -\frac{\pi}{8}$$ (quarter period): $$y = 3 \cos\left(4\left(-\frac{\pi}{8}\right) + \pi\right) = 3 \cos(-\frac{\pi}{2} + \pi) = 3 \cos\left(\frac{\pi}{2}\right) = 0$$ - At $$x = -\frac{\pi}{4} + \frac{\pi}{4} = 0$$ (half period): $$y = 3 \cos(4 \cdot 0 + \pi) = 3 \cos(\pi) = 3 \times (-1) = -3$$ - At $$x = -\frac{\pi}{4} + \frac{3\pi}{8} = \frac{\pi}{8}$$ (three-quarters period): $$y = 3 \cos\left(4 \cdot \frac{\pi}{8} + \pi\right) = 3 \cos\left(\frac{\pi}{2} + \pi\right) = 3 \cos\left(\frac{3\pi}{2}\right) = 0$$ - At $$x = -\frac{\pi}{4} + \frac{\pi}{2} = \frac{\pi}{4}$$ (end of period): $$y = 3 \cos\left(4 \cdot \frac{\pi}{4} + \pi\right) = 3 \cos(\pi + \pi) = 3 \cos(2\pi) = 3$$ 5. **Summary of key coordinates:** $$\left(-\frac{\pi}{4}, 3\right), \left(-\frac{\pi}{8}, 0\right), (0, -3), \left(\frac{\pi}{8}, 0\right), \left(\frac{\pi}{4}, 3\right)$$ 6. **Graph description:** The graph is a cosine wave with amplitude 3, period $$\frac{\pi}{2}$$, shifted left by $$\frac{\pi}{4}$$. It starts at maximum 3, crosses zero at quarter and three-quarter periods, reaches minimum -3 at half period, and returns to maximum at the end of the period.