1. **State the problem:** We are given the function $f(x) = 2x^2 - 1$ and want to understand its properties. 2. **Formula and rules:** This is a quadratic function of the form $f(x) = ax^2 + bx + c$ where $a=2$, $b=0$, and $c=-1$. 3. **Find the vertex:** The vertex of a parabola $ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Here, $x = -\frac{0}{2 \times 2} = 0$. 4. **Evaluate $f(x)$ at the vertex:** $f(0) = 2(0)^2 - 1 = -1$. So the vertex is at $(0, -1)$. 5. **Determine if the parabola opens up or down:** Since $a=2 > 0$, the parabola opens upwards, so the vertex is a minimum point. 6. **Find the y-intercept:** The y-intercept is $f(0) = -1$. 7. **Find the x-intercepts:** Solve $2x^2 - 1 = 0$. $$2x^2 = 1$$ $$x^2 = \frac{1}{2}$$ $$x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$ So the x-intercepts are at $\left(\frac{\sqrt{2}}{2}, 0\right)$ and $\left(-\frac{\sqrt{2}}{2}, 0\right)$. **Final answer:** The function $f(x) = 2x^2 - 1$ is a parabola opening upwards with vertex at $(0, -1)$, y-intercept at $(0, -1)$, and x-intercepts at $\left(\pm \frac{\sqrt{2}}{2}, 0\right)$.