The general formula is #f(c)+f'(c)(x-c)+(f''(c))/(2!)(x-c)^2+(f'''(c))/(3!)(x-c)^3+\cdots#.
Since #f(x)=\sqrt{1-x}=(1-x)^{1/2}# we get #f'(x)=-\frac{1}{2}(1-x)^{-1/2}#, #f''(x)=-\frac{1}{4}(1-x)^{-3/2}#, and #f'''(x)=-\frac{3}{8}(1-x)^{-5/2}#.
Since #c=-3#, we get #f(c)=\sqrt{4}=2#, #f'(c)=-\frac{1}{2\sqrt{4}}=-\frac{1}{4}#, #f''(c)=-\frac{1}{4\cdot 4^{3/2}}=\frac{1}{32}#, and #f'''(c)=-\frac{3}{8\cdot 4^{5/2}}=-\frac{3}{256}#.
Therefore, #\sqrt{1-x}=2-\frac{1}{4}(x+3)-\frac{1}{64}(x+3)^{2}-\frac{1}{512}(x+3)^{3}+\cdots#.
