1. **State the problem:** Graph the function $$y = 3 \cos(4x + \pi)$$ over one period and find key coordinates. 2. **Identify the period:** The general form is $$y = A \cos(Bx + C)$$ where the period $$T = \frac{2\pi}{|B|}$$. Here, $$B = 4$$, so $$T = \frac{2\pi}{4} = \frac{\pi}{2}$$. 3. **Find the phase shift:** The phase shift $$\phi = -\frac{C}{B} = -\frac{\pi}{4}$$. This means the graph is shifted left by $$\frac{\pi}{4}$$. 4. **Determine key points:** For cosine, key points in one period start at phase shift and occur every $$\frac{T}{4} = \frac{\pi}{8}$$. Calculate x-values: - Start: $$x_0 = -\frac{\pi}{4}$$ - Next points: $$x_1 = x_0 + \frac{\pi}{8} = -\frac{\pi}{8}$$ $$x_2 = x_0 + 2\cdot\frac{\pi}{8} = 0$$ $$x_3 = x_0 + 3\cdot\frac{\pi}{8} = \frac{\pi}{8}$$ $$x_4 = x_0 + 4\cdot\frac{\pi}{8} = \frac{\pi}{4}$$ 5. **Calculate y-values:** $$y = 3 \cos(4x + \pi)$$ At each x: - $$y_0 = 3 \cos(4(-\frac{\pi}{4}) + \pi) = 3 \cos(-\pi + \pi) = 3 \cos(0) = 3$$ - $$y_1 = 3 \cos(4(-\frac{\pi}{8}) + \pi) = 3 \cos(-\frac{\pi}{2} + \pi) = 3 \cos(\frac{\pi}{2}) = 0$$ - $$y_2 = 3 \cos(4(0) + \pi) = 3 \cos(\pi) = 3(-1) = -3$$ - $$y_3 = 3 \cos(4(\frac{\pi}{8}) + \pi) = 3 \cos(\frac{\pi}{2} + \pi) = 3 \cos(\frac{3\pi}{2}) = 0$$ - $$y_4 = 3 \cos(4(\frac{\pi}{4}) + \pi) = 3 \cos(\pi + \pi) = 3 \cos(2\pi) = 3$$ 6. **Summary of key points:** $$\left(-\frac{\pi}{4}, 3\right), \left(-\frac{\pi}{8}, 0\right), (0, -3), \left(\frac{\pi}{8}, 0\right), \left(\frac{\pi}{4}, 3\right)$$ 7. **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 at $$x = -\frac{\pi}{4}$$, crosses zero at $$x = -\frac{\pi}{8}$$, minimum -3 at $$x=0$$, zero at $$x=\frac{\pi}{8}$$, and back to max 3 at $$x=\frac{\pi}{4}$$. This completes one full period.