Mark M. answered 11/23/22
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f(x) = e-3x
= f(-1/3) + f'(-1/3)(x - (-1/3)) + f"(-1/3) / 2! (x - (-1/3))2 + f'''(-1/3) / 3! (x - (-1/3))3 + ...
= e - 3e(x + 1/3) + 9e / (2!) (x + 1/3)2 - 27e / 3! (x + 1/3)3 + ...
= e∑(n = 0 to ∞) [(-3)n(x + 1/3)n / n!]