Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necessary). \[ y=x^{2}+10 x-11 \] Answer Altemp, cour ers Roots: \( \square \) and \( \square \) Anets (tu)

To find the **roots** and the **vertex** of the quadratic function \( y = x^{2} + 10x - 11 \), follow these steps: ### 1. Finding the Roots The roots (also known as zeros or solutions) of the quadratic equation \( y = x^{2} + 10x - 11 \) can be found using the **quadratic formula**: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] For the given equation: - \( a = 1 \) - \( b = 10 \) - \( c = -11 \) **Step-by-Step Calculation:** 1. **Calculate the discriminant (\( D \))**: \[ D = b^{2} - 4ac = (10)^{2} - 4(1)(-11) = 100 + 44 = 144 \] 2. **Find the square root of the discriminant**: \[ \sqrt{D} = \sqrt{144} = 12 \] 3. **Calculate the two roots**: \[ x = \frac{-10 \pm 12}{2(1)} \] - **First root**: \[ x = \frac{-10 + 12}{2} = \frac{2}{2} = 1 \] - **Second root**: \[ x = \frac{-10 - 12}{2} = \frac{-22}{2} = -11 \] **Roots:** \[ x = 1 \quad \text{and} \quad x = -11 \] ### 2. Finding the Vertex The vertex of a quadratic function \( y = ax^{2} + bx + c \) is given by the point \( (h, k) \), where: \[ h = -\frac{b}{2a} \quad \text{and} \quad k = y(h) \] **Step-by-Step Calculation:** 1. **Calculate \( h \)**: \[ h = -\frac{10}{2(1)} = -5 \] 2. **Calculate \( k \)** by substituting \( h \) back into the original equation: \[ k = (-5)^{2} + 10(-5) - 11 = 25 - 50 - 11 = -36 \] **Vertex:** \[ (-5,\ -36) \] ### **Final Answer** - **Roots:** \( x = 1 \) and \( x = -11 \) - **Vertex:** \( (-5,\ -36) \)