1. The problem is to analyze the function $x^2 - 4$. 2. This is a quadratic function in the form $f(x) = ax^2 + bx + c$ where $a=1$, $b=0$, and $c=-4$. 3. Important rules for quadratics: - The graph is a parabola. - The vertex form is $f(x) = a(x-h)^2 + k$ where $(h,k)$ is the vertex. - The axis of symmetry is $x = -\frac{b}{2a}$. - The roots (x-intercepts) are found by solving $x^2 - 4 = 0$. 4. Find the roots: $$x^2 - 4 = 0$$ $$x^2 = 4$$ $$x = \pm 2$$ 5. Find the vertex: Since $b=0$, axis of symmetry is $x=0$. Evaluate $f(0) = 0^2 - 4 = -4$. So vertex is at $(0, -4)$. 6. The parabola opens upwards because $a=1 > 0$. 7. Summary: - Roots at $x=2$ and $x=-2$. - Vertex at $(0, -4)$. - Axis of symmetry $x=0$. - Opens upward. Final answer: The function $x^2 - 4$ has roots at $x=\pm 2$ and vertex at $(0,-4)$.