1. The problem involves identifying the correct inequality for the graph of a parabola opening downward with shading above the curve. 2. The general form of a downward-opening parabola is $$y = -x^2 + bx + c$$ where the coefficient of $x^2$ is negative. 3. The vertex form of a parabola is $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. Here, the vertex is approximately $(-3,0)$. 4. Using the vertex form, rewrite the parabola as $$y = -(x+3)^2 + 0$$ which expands to $$y = -x^2 - 6x - 9$$. 5. The given parabola in the problem is close to $$y = -x^2 - 6x + 8$$, which is shifted upward by 17 units compared to the vertex form expansion. 6. The inequalities given are in the form $$8 - y \leq -x^2 + 6x$$, etc. Rearranging one example: $$8 - y \leq -x^2 + 6x \implies y \geq 8 + x^2 - 6x$$ 7. Since the parabola opens downward and shading is above the curve, the correct inequality should be $$y \geq -x^2 - 6x + 8$$. 8. Rearranging this to the form given: $$8 - y \leq -x^2 - 6x$$ 9. Therefore, the correct inequality describing the graph is: $$8 - y \leq -x^2 - 6x$$ 10. This matches the option: $$8 - y \leq -x^2 - 6x$$. Final answer: $$8 - y \leq -x^2 - 6x$$