1. **State the problem:** We have the parabola given by $$y = (9 - x)(x - 3)$$ and the horizontal line $$y = k - 3$$ where $$k > 3$$. We need to find the area of the shaded region bounded by the parabola, the line, and vertical boundaries defined by points A and B. 2. **Identify points A and B:** - The parabola intersects the x-axis at points where $$y=0$$, so: $$0 = (9 - x)(x - 3)$$ which gives $$x=3$$ or $$x=9$$. Therefore, point B is at $$x=9$$. - The maximum point of the parabola occurs at the vertex. Expanding the parabola: $$y = (9 - x)(x - 3) = 9x - 27 - x^2 + 3x = -x^2 + 12x - 27$$ - The vertex $$x$$-coordinate is $$x = -\frac{b}{2a} = -\frac{12}{2(-1)} = 6$$. - The vertical line through the maximum point intercepts the x-axis at $$x=6$$, so point A is at $$x=6$$. 3. **Find the intersection point C(k, k - 3):** This point lies on both the parabola and the line: $$k - 3 = (9 - k)(k - 3)$$ - But since C is defined as $$(k, k-3)$$ and lies on $$(9-x)(x-3)$$, it's consistent. 4. **Area of the shaded region:** The shaded area lies between $$x=6$$ and $$x=k$$ (where $$6 < k < 9$$ since $$k > 3$$ and point C is on the curve). - The area $$A$$ is given by the integral of the difference between the parabola and the line from $$x=6$$ to $$x=k$$: $$A = \int_{6}^{k} \left[(9 - x)(x - 3) - (k - 3) \right] dx$$ 5. **Simplify the integrand:** $$ (9 - x)(x - 3) = -x^{2} + 12x - 27$$ $$ \Rightarrow \text{Integrand} = (-x^{2} + 12x - 27) - (k - 3) = -x^{2} + 12x - 27 - k + 3 = -x^{2} + 12x - (k + 24) $$ 6. **Calculate the integral:** $$A = \int_{6}^{k} (-x^{2} + 12x - (k + 24)) dx$$ $$= \left[-\frac{x^{3}}{3} + 6x^{2} - (k + 24)x \right]_{6}^{k}$$ 7. **Evaluate at the bounds:** - At $$x = k$$: $$-\frac{k^{3}}{3} + 6k^{2} - (k + 24)k = -\frac{k^{3}}{3} + 6k^{2} - k^{2} - 24k = -\frac{k^{3}}{3} + 5k^{2} - 24k$$ - At $$x = 6$$: $$-\frac{6^{3}}{3} + 6 \times 6^{2} - (k + 24)6 = -\frac{216}{3} + 6 \times 36 - 6k - 144 = -72 + 216 - 6k -144 = 0 - 6k = -6k$$ 8. **Calculate the definite integral:** $$A = \left(-\frac{k^{3}}{3} + 5k^{2} - 24k\right) - (-6k) = -\frac{k^{3}}{3} + 5k^{2} - 24k + 6k = -\frac{k^{3}}{3} + 5k^{2} - 18k$$ **Final answer:** $$\boxed{A = -\frac{k^{3}}{3} + 5k^{2} - 18k}$$ (valid for $$6 \leq k \leq 9$$)