Answer
$p_2(0)=f(0)$
$p'_2(0)=f'(0)$
$p"_2(0)=f"(0)$
Work Step by Step
We have to approximate $f(x)$ by a Taylor polynomial of degree $n=2$ for $a=0$.
The Taylor polynomial approximating $f(x)$ is:
$p_2(x)=f(0)+f'(0)x+\dfrac{f"(0)}{2!}x^2$
The conditions that $p_2(x)$ must check are:
$p_2(0)=f(0)$
$p'_2(0)=f'(0)$
$p"_2(0)=f"(0)$