1. **State the problem:** Find the slope of the graph of the function $$f(x) = -3e^x + x^3 - 2x$$ at the point where $$x = -2$$. 2. **Recall the formula:** The slope of the graph at a point is the value of the derivative $$f'(x)$$ at that point. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(-3e^x) + \frac{d}{dx}(x^3) - \frac{d}{dx}(2x)$$ Using derivative rules: $$f'(x) = -3e^x + 3x^2 - 2$$ 4. **Evaluate the derivative at $$x = -2$$:** $$f'(-2) = -3e^{-2} + 3(-2)^2 - 2$$ $$= -3e^{-2} + 3 \times 4 - 2$$ $$= -3e^{-2} + 12 - 2$$ $$= -3e^{-2} + 10$$ 5. **Rewrite $$e^{-2}$$ as $$\frac{1}{e^2}$$:** $$f'(-2) = -3 \times \frac{1}{e^2} + 10 = \frac{-3}{e^2} + 10$$ **Final answer:** The slope of the graph at $$x = -2$$ is $$\boxed{\frac{-3}{e^2} + 10}$$, which corresponds to option (4).