1. The problem is to analyze the function $f(x) = 9 - x^2$ and understand its graph properties. 2. This is a quadratic function in the form $f(x) = a x^2 + b x + c$ where $a = -1$, $b = 0$, and $c = 9$. 3. The parabola opens downward since $a = -1 < 0$. 4. The vertex of a parabola in standard form $a x^2 + b x + c$ is at $x = -\frac{b}{2a} = -\frac{0}{2 \times -1} = 0$. 5. Substitute $x=0$ into the function to find the vertex's y-coordinate: $f(0) = 9 - 0 = 9$. So the vertex is at $(0,9)$. 6. To find the x-intercepts, solve $0 = 9 - x^2$ which gives $x^2 = 9$ and thus $x = \pm 3$. 7. So the x-intercepts are at $(-3,0)$ and $(3,0)$. 8. The y-intercept is found by evaluating $f(0) = 9$, so $(0,9)$. 9. The parabola is symmetric about the y-axis because the function contains only even powers of $x$ and no linear term. Final answer: The function $f(x) = 9 - x^2$ is a downward opening parabola with vertex at $(0,9)$, x-intercepts at $(-3,0)$ and $(3,0)$, y-intercept at $(0,9)$, and is symmetric about the y-axis.