1. **State the problem:** Find the roots and vertex of the quadratic function $$y = -x^2 + 12x - 11$$. 2. **Formula for roots:** The roots of a quadratic $$ax^2 + bx + c = 0$$ are given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = -1$$, $$b = 12$$, and $$c = -11$$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 12^2 - 4(-1)(-11) = 144 - 44 = 100$$ 4. **Find the roots:** $$x = \frac{-12 \pm \sqrt{100}}{2(-1)} = \frac{-12 \pm 10}{-2}$$ 5. **Calculate each root:** - For the plus sign: $$x_1 = \frac{-12 + 10}{-2} = \frac{-2}{-2} = 1$$ - For the minus sign: $$x_2 = \frac{-12 - 10}{-2} = \frac{-22}{-2} = 11$$ 6. **Formula for vertex:** The vertex $$ (h, k) $$ of a parabola $$y = ax^2 + bx + c$$ is given by: $$h = -\frac{b}{2a}$$ $$k = f(h) = a h^2 + b h + c$$ 7. **Calculate vertex x-coordinate:** $$h = -\frac{12}{2(-1)} = -\frac{12}{-2} = 6$$ 8. **Calculate vertex y-coordinate:** $$k = - (6)^2 + 12(6) - 11 = -36 + 72 - 11 = 25$$ 9. **Final answers:** - Roots: $$1$$ and $$11$$ - Vertex: $$(6, 25)$$ The parabola opens downward because $$a = -1 < 0$$, so the vertex is a maximum point.