= Solution
The successive <derivatives> are $f^{(n)}(x)=(-6)^ne^{-6x}$, so $f^{(n)}(-4)=(-6)^ne^{24}$. Substitution in the <Taylor series> formula about $a=-4$ gives
$$
\boxed{e^{-6x}=e^{24}\sum_{n=0}^{\infty}\frac{(-6)^n(x+4)^n}{n!}
=e^{24}\left[1-6(x+4)+18(x+4)^2-36(x+4)^3+\cdots\right].}
$$
The <ratio test> shows convergence for every finite real or complex $x$: the absolute ratio of successive terms is $6|x+4|/(n+1)\to0$. This is also the usual <exponential function> series applied to $e^{24}e^{-6(x+4)}$, proving that the series equals the stated function everywhere.
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