1. **State the problem:** Find the coordinates of the minimum point of the curve given by the function $$y = 2x^2(3 + x)$$. 2. **Rewrite the function:** Expand the expression to simplify differentiation. $$y = 2x^2 \times 3 + 2x^2 \times x = 6x^2 + 2x^3$$ 3. **Find the derivative:** To find critical points (potential minima or maxima), differentiate $$y$$ with respect to $$x$$. $$\frac{dy}{dx} = \frac{d}{dx}(6x^2 + 2x^3) = 12x + 6x^2$$ 4. **Set the derivative equal to zero to find critical points:** $$12x + 6x^2 = 0$$ Factor out $$6x$$: $$6x(2 + x) = 0$$ 5. **Solve for $$x$$:** $$6x = 0 \Rightarrow x = 0$$ $$2 + x = 0 \Rightarrow x = -2$$ 6. **Determine the nature of each critical point using the second derivative:** Second derivative: $$\frac{d^2y}{dx^2} = \frac{d}{dx}(12x + 6x^2) = 12 + 12x$$ Evaluate at $$x=0$$: $$12 + 12 \times 0 = 12 > 0$$ (minimum) Evaluate at $$x=-2$$: $$12 + 12 \times (-2) = 12 - 24 = -12 < 0$$ (maximum) 7. **Find the $$y$$-coordinate of the minimum point at $$x=0$$:** $$y = 6(0)^2 + 2(0)^3 = 0$$ 8. **Conclusion:** The minimum point is at $$(0, 0)$$. **Final answer:** A (0, 0)