Let \(f\) be a function with \(f(3)=2, f'(3)=-1, f''(3)=6, \text{ and } f'''(3)=12 \). Which of the following is the third-degree Taylor polynomial for \(f\) about \(x=3\) ?

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We can use the equation:

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

When \(c=3, n=3\):

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

If you feel more comfortable with the other method (see previous question's explanation), you can take the first, second, and third derivative of each answer choice and evaluate at \(x=3\).