Find the Taylor polynomial $T_{3}$ for the function $f(x)=e^{-x}$ at the number $a=-4$ A. $e^{4}+e^{4}(x-4)+\frac{e^{4}}{2}(x-4)^{2}+\frac{e^{4}}{3 !}(x-4)^{3}$ B. $e^{4}-e^{4}(x+4)+\frac{e^{4}}{2}(x+4)^{2}-\frac{e^{4}}{3 i}(x+4)^{3}$ C. $e^{4}-e^{4}(x-4)+\frac{e^{4}}{2}(x-4)^{2}-\frac{e^{4}}{3 !}(x-4)^{3}$ D. $e^{4}-e^{4}(x+4)+\frac{e^{4}}{2}(x+4)^{2}-\frac{e^{4}}{3}(x+4)^{3}$ E. $e^{4}+e^{4}(x+4)+\frac{e^{4}}{2}(x+4)^{2}+\frac{e^{4}}{3 !}(x+4)^{3}$ F. none of these

Step 1 ：The Taylor polynomial of degree 3 for a function \(f(x)\) at a number \(a\) is given by: \[T_{3}(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!}\]

Step 2 ：Given \(f(x) = e^{-x}\), we first need to calculate its derivatives at \(a = -4\).

Step 3 ：The first derivative of \(f(x)\) is \(f'(x) = -e^{-x}\).

Step 4 ：The second derivative of \(f(x)\) is \(f''(x) = e^{-x}\).

Step 5 ：The third derivative of \(f(x)\) is \(f'''(x) = -e^{-x}\).

Step 6 ：At \(a = -4\), we have: \(f(-4) = e^4\), \(f'(-4) = -e^4\), \(f''(-4) = e^4\), and \(f'''(-4) = -e^4\).

Step 7 ：Substituting these values into the Taylor polynomial formula, we get: \[T_{3}(x) = e^4 - e^4(x+4) + \frac{e^4(x+4)^2}{2!} - \frac{e^4(x+4)^3}{3!}\]

Step 8 ：This corresponds to option B. \(\boxed{\text{Option B}}\)