1. **State the problem:** We have a parabola representing the ball's height: $$y = -x^2 + 6x$$ and a horizontal line at height $$y = 8$$. We want to find where the ball reaches this height and discuss the two possible times it is at this height. 2. **Set up the equation:** To find the intersection points, set the parabola equal to the line: $$-x^2 + 6x = 8$$ 3. **Rearrange the equation:** Move all terms to one side: $$-x^2 + 6x - 8 = 0$$ Multiply both sides by -1 to simplify: $$x^2 - 6x + 8 = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-6$$, and $$c=8$$. Calculate the discriminant: $$\Delta = (-6)^2 - 4 \times 1 \times 8 = 36 - 32 = 4$$ Calculate the roots: $$x = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2}$$ So, $$x_1 = \frac{6 + 2}{2} = 4$$ $$x_2 = \frac{6 - 2}{2} = 2$$ 5. **Interpretation:** The ball reaches the height of 8 at $$x=2$$ and $$x=4$$. 6. **Discuss the two times:** These two values represent the times when the ball is at height 8. At $$x=2$$, the ball is ascending to that height, and at $$x=4$$, it is descending back down. **Final answer:** The ball reaches the height of 8 at $$x=2$$ and $$x=4$$.