The problem asks us to find the orientation, vertex, y-intercept, and axis of symmetry of the parabola defined by the equation $y = 2x^2 - 2$.

1. Problem Description

The problem asks us to find the orientation, vertex, y-intercept, and axis of symmetry of the parabola defined by the equation $y = 2x^2 - 2$.

2. Solution Steps

First, let's determine the orientation of the parabola. The coefficient of the $x^2$ term is 2, which is positive. Therefore, the parabola opens upwards.

Next, we find the vertex of the parabola. The equation is in the form $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex. In our case, we have $y = 2(x-0)^2 - 2$. So, $h = 0$ and $k = -2$. Therefore, the vertex is $(0, -2)$.

Now, we find the y-intercept. The y-intercept occurs when $x = 0$. Substituting $x=0$ into the equation $y = 2x^2 - 2$, we get $y = 2(0)^2 - 2 = -2$. Thus, the y-intercept is $(0, -2)$.

Finally, we find the axis of symmetry. The axis of symmetry is a vertical line that passes through the vertex of the parabola. Since the x-coordinate of the vertex is 0, the equation of the axis of symmetry is $x = 0$.

3. Final Answer

Orientation: Opens up

Vertex (x, y): (0, -2)

y-intercept (x, y): (0, -2)

Axis of symmetry: x = 0