To find the Taylor polynomial of order 3 generated by $f(x) = x^2$ at $a = 3$, we need to use the formula for the Taylor series expansion:

$P_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3$

where $n = 3$ in this case.

1. Compute $f(a)$: $f(3) = 3^2 = 9$

2. Compute the first derivative $f'(x)$: $f'(x) = 2x$ $f'(3) = 2 \cdot 3 = 6$

3. Compute the second derivative $f''(x)$: $f''(x) = 2$ $f''(3) = 2$

4. Compute the third derivative $f'''(x)$: $f'''(x) = 0$ $f'''(3) = 0$

Now, substitute these values into the Taylor series formula:

$P_3(x) = f(3) + f'(3)(x - 3) + \frac{f''(3)}{2!}(x - 3)^2 + \frac{f'''(3)}{3!}(x - 3)^3$

$P_3(x) = 9 + 6(x - 3) + \frac{2}{2}(x - 3)^2 + \frac{0}{6}(x - 3)^3$

$P_3(x) = 9 + 6(x - 3) + (x - 3)^2$

Thus, the correct Taylor polynomial of order 3 for $f(x) = x^2$ at $a = 3$ is:

$P_3(x) = 9 + 6(x - 3) + (x - 3)^2$

Therefore, the correct choice is: $\boxed{P_3(x) = 9 + 6(x - 3) + (x - 3)^2}$

Would you like more details or have any questions?

Here are some additional questions to deepen your understanding:

1. How do you derive the formula for the Taylor series expansion?
2. Why is the third derivative $f'''(x)$ zero in this case?
3. What is the significance of each term in the Taylor polynomial?
4. How does the accuracy of a Taylor polynomial change with the order $n$?
5. Can you explain the process of finding higher-order derivatives for more complex functions?
6. What are some practical applications of Taylor series in real-world scenarios?
7. How do you find the radius of convergence for a Taylor series?
8. What happens to the Taylor series as $x$ approaches $a$?

Tip: Always verify each term in the Taylor polynomial by computing the necessary derivatives carefully, ensuring all calculations align with the given function and point of expansion.