1. **State the problem:** We want to graph the cubic function $$A(x) = \frac{1}{2} x^3 - \frac{9}{2} x^2 + 9 x$$ defined on the interval $$[0,3]$$. 2. **Understand the function:** This is a cubic polynomial. Cubic functions can have up to two turning points (local maxima or minima) and one inflection point. 3. **Find critical points:** To find turning points, compute the derivative: $$A'(x) = \frac{d}{dx} \left( \frac{1}{2} x^3 - \frac{9}{2} x^2 + 9 x \right) = \frac{3}{2} x^2 - 9 x + 9$$ 4. **Set derivative to zero to find critical points:** $$\frac{3}{2} x^2 - 9 x + 9 = 0$$ Multiply both sides by 2 to clear fraction: $$3 x^2 - 18 x + 18 = 0$$ Divide both sides by 3: $$\cancel{3} x^2 - \cancel{18} x + \cancel{18} = 0 \Rightarrow x^2 - 6 x + 6 = 0$$ 5. **Solve quadratic equation:** $$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2} = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2 \sqrt{3}}{2} = 3 \pm \sqrt{3}$$ 6. **Check if critical points lie in domain [0,3]:** - $$3 - \sqrt{3} \approx 3 - 1.732 = 1.268$$ (inside domain) - $$3 + \sqrt{3} \approx 3 + 1.732 = 4.732$$ (outside domain) Only $$x = 3 - \sqrt{3}$$ is relevant. 7. **Find function values at critical points and endpoints:** - At $$x=0$$: $$A(0) = 0$$ - At $$x=3 - \sqrt{3}$$: $$A(3 - \sqrt{3}) = \frac{1}{2} (3 - \sqrt{3})^3 - \frac{9}{2} (3 - \sqrt{3})^2 + 9 (3 - \sqrt{3})$$ - At $$x=3$$: $$A(3) = \frac{1}{2} (3)^3 - \frac{9}{2} (3)^2 + 9 (3) = \frac{1}{2} \cdot 27 - \frac{9}{2} \cdot 9 + 27 = 13.5 - 40.5 + 27 = 0$$ 8. **Summary:** The graph starts at (0,0), rises to a local maximum near $$x=1.268$$, then decreases back to (3,0). The curve is smooth and continuous. 9. **Graphing:** Plot the function $$A(x)$$ on $$[0,3]$$ showing the turning point at $$x=3 - \sqrt{3}$$. Final answer: The curve of $$A(x) = \frac{1}{2} x^3 - \frac{9}{2} x^2 + 9 x$$ on $$[0,3]$$ starts at zero, rises to a local maximum near $$x=1.268$$, then falls back to zero at $$x=3$$.