(1 point) For the quadratic function $f(x)=2 x^{2}+2 x-1$, find the vertex, the axis of symmetry, and the $x$-intercept(s). Express the vertex as a coordinate pair. If there is more than one intercept, enter as a comma separated list of points. - vertex: $\square$ help (points) - axis of symmetry: ? $\hat{\approx} \square$ help (numbers) - $x$-intercept(s): $\square$ help (points) Note: You can earn partial credit on this problem.

Step 1 ：The vertex of a quadratic function is given by the formula \((-\frac{b}{2a}, f(-\frac{b}{2a}))\). Here, \(a = 2\) and \(b = 2\). So, the x-coordinate of the vertex is \(-\frac{b}{2a} = -\frac{2}{2*2} = -\frac{1}{2}\).

Step 2 ：Substitute \(x = -\frac{1}{2}\) into the function to find the y-coordinate of the vertex: \(f(-\frac{1}{2}) = 2(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 1 = 2*\frac{1}{4} - 1 - 1 = -\frac{1}{2}\).

Step 3 ：So, the vertex of the function is \((-\frac{1}{2}, -\frac{1}{2})\).

Step 4 ：The axis of symmetry of a quadratic function is the vertical line \(x = -\frac{b}{2a}\). So, the axis of symmetry here is \(x = -\frac{1}{2}\).

Step 5 ：The x-intercepts of the function are the solutions to the equation \(f(x) = 0\), or \(2x^2 + 2x - 1 = 0\). This can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 6 ：Here, \(a = 2\), \(b = 2\), and \(c = -1\). So, the solutions are: \(x = \frac{-2 \pm \sqrt{2^2 - 4*2*(-1)}}{2*2} = \frac{-2 \pm \sqrt{4 + 8}}{4} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-2 \pm 2\sqrt{3}}{4} = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}\).

Step 7 ：So, the x-intercepts of the function are \((-\frac{1}{2} - \frac{\sqrt{3}}{2}, 0)\) and \((-\frac{1}{2} + \frac{\sqrt{3}}{2}, 0)\).

Step 8 ：In conclusion: \(\boxed{\text{The vertex is } (-\frac{1}{2}, -\frac{1}{2})}\), \(\boxed{\text{The axis of symmetry is } x = -\frac{1}{2}}\), \(\boxed{\text{The x-intercepts are } (-\frac{1}{2} - \frac{\sqrt{3}}{2}, 0) \text{ and } (-\frac{1}{2} + \frac{\sqrt{3}}{2}, 0)}\).