1. The problem is to graph the quadratic function $f(x) = -\frac{1}{2}x^2 + 3$. 2. This is a quadratic function, which generally has the form $f(x) = ax^2 + bx + c$. 3. Here, $a = -\frac{1}{2}$, $b = 0$, and $c = 3$. Since $a$ is negative, the parabola opens downward. 4. The vertex form of a quadratic is $f(x) = a(x-h)^2 + k$, where $(h,k)$ is the vertex. 5. Since $b=0$, the vertex is at $x = -\frac{b}{2a} = 0$. 6. Substitute $x=0$ into the function to find $k$: $f(0) = -\frac{1}{2}(0)^2 + 3 = 3$. 7. So the vertex is at $(0,3)$, the highest point of the parabola. 8. The parabola opens downward because $a < 0$. 9. To find the x-intercepts, set $f(x) = 0$: $$0 = -\frac{1}{2}x^2 + 3$$ Multiply both sides by $-2$ to clear the fraction: $$0 = x^2 - 6$$ Intermediate step with cancellation: $$0 = \cancel{-2} \times 0 = \cancel{-2} \times \left(-\frac{1}{2}x^2 + 3\right) = x^2 - 6$$ 10. Solve for $x^2$: $$x^2 = 6$$ 11. Take the square root: $$x = \pm \sqrt{6}$$ 12. The y-intercept is found by evaluating $f(0)$, which is $3$. 13. The graph is a downward-opening parabola with vertex at $(0,3)$ and x-intercepts at $\left(-\sqrt{6}, 0\right)$ and $\left(\sqrt{6}, 0\right)$. Final answer: The graph of $f(x) = -\frac{1}{2}x^2 + 3$ is a downward-opening parabola with vertex at $(0,3)$ and x-intercepts at $\pm \sqrt{6}$.