Roots Vertex 6Faf86

1. **State the problem:** Find the roots and vertex of the quadratic function $$y = -x^2 - 10x + 11$$. 2. **Formula for roots:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = -1$$, $$b = -10$$, and $$c = 11$$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-10)^2 - 4(-1)(11) = 100 + 44 = 144$$ 4. **Calculate the roots:** $$x = \frac{-(-10) \pm \sqrt{144}}{2(-1)} = \frac{10 \pm 12}{-2}$$ 5. **Find each root:** - For the plus sign: $$x = \frac{10 + 12}{-2} = \frac{22}{-2} = -11$$ - For the minus sign: $$x = \frac{10 - 12}{-2} = \frac{-2}{-2} = 1$$ 6. **Roots:** $$x = -11$$ and $$x = 1$$ 7. **Formula for vertex:** The vertex $$x$$-coordinate is $$x = -\frac{b}{2a}$$. 8. **Calculate vertex $$x$$-coordinate:** $$x = -\frac{-10}{2(-1)} = \frac{10}{-2} = -5$$ 9. **Calculate vertex $$y$$-coordinate:** Substitute $$x = -5$$ into the original equation: $$y = -(-5)^2 - 10(-5) + 11 = -25 + 50 + 11 = 36$$ 10. **Vertex:** $$(-5, 36)$$ **Final answer:** Roots: $$-11$$ and $$1$$ Vertex: $$(-5, 36)$$