1. **State the problem:** Graph the function $f(x) = -x^2 + 8$ and plot the vertex and another point on the parabola. 2. **Formula and rules:** This is a quadratic function in the form $f(x) = ax^2 + bx + c$ where $a = -1$, $b = 0$, and $c = 8$. Since $a < 0$, the parabola opens downward. 3. **Find the vertex:** The vertex of a parabola $f(x) = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Here, $x = -\frac{0}{2(-1)} = 0$. 4. **Calculate the vertex's y-coordinate:** Substitute $x=0$ into $f(x)$: $$f(0) = -0^2 + 8 = 8$$ So the vertex is at $(0, 8)$. 5. **Plot another point:** Choose $x=1$: $$f(1) = -(1)^2 + 8 = -1 + 8 = 7$$ So another point on the parabola is $(1, 7)$. 6. **Plot symmetric point:** Because the parabola is symmetric about the vertex, the point $(-1, 7)$ is also on the parabola. 7. **Summary:** The parabola opens downward with vertex at $(0, 8)$ and passes through points $(1, 7)$ and $(-1, 7)$. This completes the graphing of $f(x) = -x^2 + 8$ with the vertex and another point plotted.