Question Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necescary). \[ y=3 x^{2}+9 x-293 \] Answer Attemptroutof2 Roots: \( \square \) and \( \square \)

To find the **roots** and the **vertex** of the quadratic function \( y = 3x^2 + 9x - 293 \), follow these steps: ### 1. Finding the Roots The roots of the quadratic equation \( 3x^2 + 9x - 293 = 0 \) can be found using the **quadratic formula**: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3 \), \( b = 9 \), and \( c = -293 \). **Step-by-Step Calculation:** 1. **Calculate the Discriminant (\( D \)):** \[ D = b^2 - 4ac = 9^2 - 4(3)(-293) = 81 + 3516 = 3597 \] 2. **Compute the Square Root of the Discriminant:** \[ \sqrt{3597} \approx 59.975 \] 3. **Find the Roots:** \[ x = \frac{-9 \pm 59.975}{2 \times 3} = \frac{-9 \pm 59.975}{6} \] - **First Root:** \[ x_1 = \frac{-9 + 59.975}{6} = \frac{50.975}{6} \approx 8.496 \] - **Second Root:** \[ x_2 = \frac{-9 - 59.975}{6} = \frac{-68.975}{6} \approx -11.496 \] ### 2. Finding the Vertex The vertex of a quadratic function \( y = ax^2 + bx + c \) is given by: \[ \left( -\frac{b}{2a}, \ y\left(-\frac{b}{2a}\right) \right) \] **Step-by-Step Calculation:** 1. **Calculate the x-coordinate of the Vertex:** \[ x_v = -\frac{9}{2 \times 3} = -\frac{9}{6} = -1.5 \] 2. **Calculate the y-coordinate by Substituting \( x_v \) into the Original Equation:** \[ y_v = 3(-1.5)^2 + 9(-1.5) - 293 = 3(2.25) - 13.5 - 293 = 6.75 - 13.5 - 293 = -299.75 \] \[ y_v \approx -299.750 \] ### **Final Answers:** - **Roots:** \( 8.496 \) and \( -11.496 \) - **Vertex:** \( (-1.500, -299.750) \)