1. Problem: Find the zeroes, domain, range, axis of symmetry, and vertex for the quadratic function $y = x^2 - 9$. 2. Formula and rules: - Zeroes are found by solving $y=0$. - Domain of any quadratic function is all real numbers: $(-\infty, \infty)$. - Range depends on the vertex and whether parabola opens up or down. - Axis of symmetry formula: $x = -\frac{b}{2a}$ for $ax^2 + bx + c$. - Vertex coordinates: $(h, k)$ where $h$ is axis of symmetry and $k = f(h)$. 3. Find zeroes: Set $y=0$: $$x^2 - 9 = 0$$ $$x^2 = 9$$ $$x = \pm 3$$ Zeroes are $x=3$ and $x=-3$. 4. Domain: All real numbers: $(-\infty, \infty)$. 5. Axis of symmetry: Here $a=1$, $b=0$, so $$x = -\frac{0}{2 \times 1} = 0$$ 6. Vertex: Calculate $y$ at $x=0$: $$y = 0^2 - 9 = -9$$ Vertex is at $(0, -9)$. 7. Range: Since $a=1 > 0$, parabola opens upward, so minimum value is vertex $y=-9$. Range is $[-9, \infty)$. Final answer: - Zeroes: $x=3, -3$ - Domain: $(-\infty, \infty)$ - Range: $[-9, \infty)$ - Axis of symmetry: $x=0$ - Vertex: $(0, -9)$