Answer
$1-x+\dfrac{x^2}{2}$
Work Step by Step
The Taylor approximation for degree $n$ centered at point $a$ can be written as:
$P_n(x)=f(a)+f'(a)(x-a)+\dfrac{1}{2}f''(a)(x-a)^2+.......+\dfrac{1}{n!}f^n(a)(x-a)^n ~~~~........(1)$----
We have: $f(x)=e^{-x} \implies f(0)=1$ with $n=2$ and $a=0$
Further, $f'(x)=-e^{-x} \implies f'(0)=-1 \\ f''(x)=e^{-x} \implies f''(0)=1$
Now, plug these values in the equation (1) to obtain:
$P_2(x)=f(0)+f'(0)\dfrac{x}{1!}+\dfrac{1}{2!}f''(0)x^2\\=1+(-1) x+\dfrac{x^2}{2!}\\=1-x+\dfrac{x^2}{2}$
