1. **State the problem:** We need to trace the polar function $$r = f(\theta) = -3 - 3 \cos(\theta)$$ starting at $$\theta = \frac{3\pi}{2}$$ and ending at $$\theta = 2\pi$$. 2. **Recall the formula:** The polar function is given by $$r = -3 - 3 \cos(\theta)$$ where $$\theta$$ ranges from $$0$$ to $$2\pi$$. 3. **Evaluate the function at key points between $$\frac{3\pi}{2}$$ and $$2\pi$$:** - At $$\theta = \frac{3\pi}{2}$$: $$r = -3 - 3 \cos\left(\frac{3\pi}{2}\right) = -3 - 3 \times 0 = -3$$ - At $$\theta = \frac{7\pi}{4}$$: $$r = -3 - 3 \cos\left(\frac{7\pi}{4}\right) = -3 - 3 \times \frac{\sqrt{2}}{2} = -3 - \frac{3\sqrt{2}}{2}$$ - At $$\theta = 2\pi$$: $$r = -3 - 3 \cos(2\pi) = -3 - 3 \times 1 = -6$$ 4. **Interpretation:** Since $$r$$ is negative in this interval, the points are plotted in the opposite direction of the angle $$\theta$$. 5. **Trace the curve:** Starting at $$\theta = \frac{3\pi}{2}$$ with $$r = -3$$, move through intermediate points like $$\frac{7\pi}{4}$$ with $$r = -3 - \frac{3\sqrt{2}}{2}$$, and end at $$\theta = 2\pi$$ with $$r = -6$$. 6. **Summary:** The trace from $$\theta = \frac{3\pi}{2}$$ to $$2\pi$$ follows the cardioid shape in the polar plot, moving through negative radius values which reflect points opposite the angle direction. **Final answer:** The portion of the curve from $$\theta = \frac{3\pi}{2}$$ to $$2\pi$$ is traced by plotting $$r = -3 - 3 \cos(\theta)$$ for $$\theta$$ in $$\left[\frac{3\pi}{2}, 2\pi\right]$$, noting that negative $$r$$ values plot points opposite to $$\theta$$.