Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necessary). \[ y=-x^{2}+10 x+24 \] Answer Attempts out of 2 Roots: \( \square \) and Vertex:

To find the roots and the vertex of the quadratic function \( y = -x^{2} + 10x + 24 \), follow these steps: ### **1. Finding the Roots:** Set \( y = 0 \) and solve for \( x \): \[ -x^{2} + 10x + 24 = 0 \] Rearrange the equation: \[ x^{2} - 10x - 24 = 0 \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -10 \), and \( c = -24 \): \[ x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-24)}}{2(1)} = \frac{10 \pm \sqrt{100 + 96}}{2} = \frac{10 \pm \sqrt{196}}{2} = \frac{10 \pm 14}{2} \] So, the roots are: \[ x = \frac{10 + 14}{2} = 12 \quad \text{and} \quad x = \frac{10 - 14}{2} = -2 \] ### **2. Finding the Vertex:** The vertex of a quadratic function \( y = ax^2 + bx + c \) is located at \( x = \frac{-b}{2a} \). Here, \( a = -1 \) and \( b = 10 \): \[ x = \frac{-10}{2(-1)} = \frac{-10}{-2} = 5 \] Now, substitute \( x = 5 \) back into the original equation to find the \( y \)-coordinate: \[ y = -(5)^2 + 10(5) + 24 = -25 + 50 + 24 = 49 \] So, the vertex is at \( (5, 49) \). ### **Final Answer:** **Roots:** \( -2 \) and \( 12 \) **Vertex:** \( (5,\ 49) \)