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 vertex: - $h = -\frac{0}{2 \times 1} = 0$ - $k = f(0) = 0^2 - 4 = -4$ - So vertex is at $(0, -4)$. 5. Find the roots: - Solve $x^2 - 4 = 0$ - $x^2 = 4$ - $x = \pm 2$ 6. The y-intercept is $f(0) = -4$. 7. Summary: - Vertex at $(0, -4)$ - Roots at $x = -2$ and $x = 2$ - Parabola opens upwards because $a=1 > 0$. Final answer: The function $x^2 - 4$ has roots at $x = -2$ and $x = 2$, vertex at $(0, -4)$, and opens upward.