To solve this problem, let's analyze the given conditions and find the graph that satisfies them. The conditions are:

Key Conditions:

1. $f'(6) = 0$: The derivative is zero at $x = 6$, indicating a critical point.
2. $f'(x) < 0$ when $x < 6$: The function is decreasing for $x < 6$.
3. $f'(x) > 0$ when $x > 6$: The function is increasing for $x > 6$.
4. $f''(3) = 0$: The second derivative is zero at $x = 3$, indicating a possible inflection point.
5. $f''(8) = 0$: The second derivative is zero at $x = 8$, indicating a possible inflection point.
6. $f''(x) < 0$ when $x < 3$ or $x > 8$: The function is concave down in these regions.
7. $f''(x) > 0$ for $3 < x < 8$: The function is concave up in this interval.
8. $\lim_{x \to \infty} f(x) = 3$ and $\lim_{x \to -\infty} f(x) = 3$: The function approaches a horizontal asymptote at $y = 3$.

Analysis:

1. At $x = 6$, there is a critical point where the function transitions from decreasing ($f'(x) < 0$) to increasing ($f'(x) > 0$). This suggests a local minimum at $x = 6$.
2. $f''(x)$ indicates changes in concavity:
   • Concave down ($f''(x) < 0$) for $x < 3$ and $x > 8$: These regions suggest the graph has downward curvature.
   • Concave up ($f''(x) > 0$) for $3 < x < 8$: This suggests upward curvature between these points.
3. Horizontal asymptotes at $y = 3$ as $x \to \pm \infty$: The function levels off to the line $y = 3$ as $x$ becomes very large or very small.
4. Points of inflection occur at $x = 3$ and $x = 8$ because the second derivative changes sign.

Matching the Graph:

Based on the conditions:

• The graph must decrease to $x = 6$, reach a minimum, then increase.
• The curvature changes from concave down to concave up at $x = 3$ and back to concave down at $x = 8$.
• The function approaches $y = 3$ asymptotically as $x \to \pm \infty$.

From the given graphs, the correct option is the one satisfying these conditions. Would you like me to analyze each graph or assist with further clarifications?