1. The problem is to analyze the function $f(x) = x^2 - 4$. 2. This is a quadratic function in the form $ax^2 + bx + c$ 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: - Since $b=0$, axis of symmetry is $x=0$. - Evaluate $f(0) = 0^2 - 4 = -4$. - Vertex is at $(0, -4)$. 5. Find the roots: - Solve $x^2 - 4 = 0$. - Add 4 to both sides: $x^2 = 4$. - Take square root: $x = \pm 2$. - Roots are $x=2$ and $x=-2$. 6. The y-intercept is found by evaluating $f(0) = -4$. 7. Summary: - Vertex: $(0, -4)$ - Roots: $x = -2, 2$ - Axis of symmetry: $x=0$ - Opens upward since $a=1 > 0$. Final answer: The parabola $y = x^2 - 4$ has vertex at $(0,-4)$, roots at $x=\pm 2$, and opens upward.