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