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 coordinates of the two \( x \)-intercepts of the parabola given by the equation \( f(x) = 2x^2 + 14x + 9 \), we need to set \( f(x) = 0 \) and solve for \( x \). Given: \( f(x) = 2x^2 + 14x + 9 \) To find the \( x \)-intercepts, we set \( f(x) = 0 \): \( 2x^2 + 14x + 9 = 0 \) Now, we can solve this quadratic equation to find the values of \( x \) where the parabola intersects the \( x \)-axis. Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(2x^{2}+14x+9=0\) - step1: Solve using the quadratic formula: \(x=\frac{-14\pm \sqrt{14^{2}-4\times 2\times 9}}{2\times 2}\) - step2: Simplify the expression: \(x=\frac{-14\pm \sqrt{14^{2}-4\times 2\times 9}}{4}\) - step3: Simplify the expression: \(x=\frac{-14\pm \sqrt{124}}{4}\) - step4: Simplify the expression: \(x=\frac{-14\pm 2\sqrt{31}}{4}\) - step5: Separate into possible cases: \(\begin{align}&x=\frac{-14+2\sqrt{31}}{4}\\&x=\frac{-14-2\sqrt{31}}{4}\end{align}\) - step6: Simplify the expression: \(\begin{align}&x=\frac{-7+\sqrt{31}}{2}\\&x=\frac{-14-2\sqrt{31}}{4}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{-7+\sqrt{31}}{2}\\&x=-\frac{7+\sqrt{31}}{2}\end{align}\) - step8: Rewrite: \(x_{1}=-\frac{7+\sqrt{31}}{2},x_{2}=\frac{-7+\sqrt{31}}{2}\) The solutions to the quadratic equation \(2x^2 + 14x + 9 = 0\) are: \( x_1 \approx -6.283882 \) and \( x_2 \approx -0.716118 \). Therefore, the coordinates of the two \( x \)-intercepts of the parabola are approximately \((-6.28, 0)\) and \((-0.72, 0)\).