1. **State the problem:** We are given the function $f(x) = x^2 - 4$ and need to analyze it. 2. **Formula and rules:** This is a quadratic function of the form $f(x) = ax^2 + bx + c$ where $a=1$, $b=0$, and $c=-4$. 3. **Find the roots (x-intercepts):** Set $f(x) = 0$: $$x^2 - 4 = 0$$ $$x^2 = 4$$ $$x = \pm 2$$ So the roots are $x=2$ and $x=-2$. 4. **Find the vertex (extremum):** The vertex of $f(x) = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Here, $x = -\frac{0}{2 \times 1} = 0$. Calculate $f(0)$: $$f(0) = 0^2 - 4 = -4$$ So the vertex is at $(0, -4)$, which is a minimum point since $a=1 > 0$. 5. **Summary:** The parabola opens upwards, has roots at $x=\pm 2$, and a minimum vertex at $(0, -4)$. **Final answer:** The function $f(x) = x^2 - 4$ has roots at $x=2$ and $x=-2$, and a minimum value of $-4$ at $x=0$.