1. **State the problem:** We need to analyze the function $f(x) = x^2 - 2$. 2. **Recall the formula and rules:** This is a quadratic function in the form $f(x) = ax^2 + bx + c$ where $a=1$, $b=0$, and $c=-2$. 3. **Find the vertex:** The vertex of a parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Here, $x = -\frac{0}{2 \times 1} = 0$. 4. **Calculate the vertex's y-coordinate:** Substitute $x=0$ into $f(x)$: $$f(0) = 0^2 - 2 = -2$$ 5. **Determine the vertex:** The vertex is at $(0, -2)$. 6. **Find the y-intercept:** The y-intercept is the value of $f(0)$, which is $-2$. 7. **Find the x-intercepts:** Solve $f(x) = 0$: $$x^2 - 2 = 0$$ $$x^2 = 2$$ $$x = \pm \sqrt{2}$$ 8. **Summary:** The parabola opens upwards (since $a=1>0$), vertex at $(0,-2)$, y-intercept at $(0,-2)$, and x-intercepts at $(\sqrt{2},0)$ and $(-\sqrt{2},0)$. **Final answer:** Vertex: $(0,-2)$, x-intercepts: $\pm \sqrt{2}$, y-intercept: $-2$.