1. The problem is to write the Taylor series formula for a function $f(x)$ centered at $x=-2$ up to the 4th derivative term. 2. The Taylor series expansion of a function $f(x)$ about a point $a$ is given by: $$f(x) = \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} (x - a)^n$$ where $f^{(n)}(a)$ is the $n$th derivative of $f$ evaluated at $x=a$, and $n!$ is the factorial of $n$. 3. In this problem, the center is $a = -2$, and the sum goes from $n=0$ to $4$. 4. Substituting $a = -2$ into the formula, we get: $$f(x) = \sum_{n=0}^4 \frac{f^{(n)}(-2)}{n!} (x - (-2))^n = \sum_{n=0}^4 \frac{f^{(n)}(-2)}{n!} (x + 2)^n$$ 5. This matches the given formula: $$f(x) = \sum_{n=0}^4 \frac{f^{(n)}(-2)}{n!} (x + 2)^n$$ 6. Note: The original formula in the prompt has $( -2 )^n$ which seems to be a misinterpretation. The correct Taylor series uses $(x - a)^n$, which here is $(x + 2)^n$. Final answer: $$f(x) = \sum_{n=0}^4 \frac{f^{(n)}(-2)}{n!} (x + 2)^n$$