Consider the parabola given by the equation: \( f(x)=2 x^{2}+14 x+9 \) Find the following for this parabola: A) The vertex: \( \left(-\frac{7}{2},-\frac{31}{2}\right) \) B) The vertical intercept is the point \( (0,9) \) C) Find the coordinates of the two \( x \)-intercepts of the parabola and write them as a list, separated by commas: It is OK to round your value(s) to to two decimal olaces.

To find the \( x \)-intercepts of the parabola given by the equation \( f(x)=2x^{2}+14x+9 \), we need to set \( f(x) = 0 \) and solve for \( x \). Start by rewriting the equation: \[ 2x^{2} + 14x + 9 = 0 \] Now, apply the quadratic formula, \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 2 \), \( b = 14 \), and \( c = 9 \). Calculate the discriminant: \[ b^{2} - 4ac = 14^{2} - 4(2)(9) = 196 - 72 = 124 \] Now, plug in the values into the quadratic formula: \[ x = \frac{-14 \pm \sqrt{124}}{2(2)} \] Simplifying \( \sqrt{124} \): \[ \sqrt{124} = \sqrt{4 \cdot 31} = 2\sqrt{31} \] Now substituting back: \[ x = \frac{-14 \pm 2\sqrt{31}}{4} = \frac{-7 \pm \sqrt{31}}{2} \] This gives us two \( x \)-intercepts: 1. \( x_1 = \frac{-7 + \sqrt{31}}{2} \) 2. \( x_2 = \frac{-7 - \sqrt{31}}{2} \) Now rounding these values to two decimal places: - \( x_1 \approx 0.29 \) - \( x_2 \approx -7.29 \) So the \( x \)-intercepts of the parabola are approximately: \( 0.29, -7.29 \)