Basis for polynomials of degree 2 or less (derivatives) I want to find a basis $\mathbb{B}$ for $P_2$ such that
${p}_\mathbb{B}$ =  $
    \begin{bmatrix} 
    p(0) \\ p'(0) \\ p''(2)/2 
\end {bmatrix}$
are the coordinates for a second degree polynome $p$.
Since $p(0) = a$, $p'(0) = b$ and $p''(0)/2 = c$, I wonder if my basis is correct: (I assumed $P_2 = a+bx+cx^2$)
$\mathbb{B} =  
    \begin{bmatrix} 
    1 \\ 0 \\ 0 
\end {bmatrix} \begin{bmatrix} 
    0 \\ 1 \\ 0 
\end {bmatrix} \begin{bmatrix} 
    0 \\ 0 \\ 1 
\end {bmatrix}$
Because $a*\begin{bmatrix} 
    1 \\ 0 \\ 0 
\end {bmatrix} +b\begin{bmatrix} 
    0 \\ 1 \\ 0 
\end {bmatrix} + c\begin{bmatrix} 
    0 \\ 0 \\ 1 
\end {bmatrix} = \begin{bmatrix} 
    a \\ b \\ c \end {bmatrix}$ which indeed is the result that I wanted. The derivates are kind of confusing though, I'm not sure if I've done this correctly.
 A: Edit: Thanks for awarding me with the best answer, but I think the explanation was not perfect, so I found a better way. If you still like the old explanation better, then you can take a look at the "Edit History".
Our goal is it to find a basis of $P_2$ such that $$p_\mathbb{B} = \begin{bmatrix} p(0) \\ p'(0) \\ p''(0)/2 \end{bmatrix} $$
for every element $p$ in $P_2$. 
Let $p = a + bx + cx^2$ an arbitrary element with coefficients $a,b,c$.
Since you have already shown us that $p(0)=a$, $p'(0) = b$ and $p''(0)/2 = c$, this means that we must find a basis $\mathbb{B}$ such that $$p_\mathbb{B} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}. $$
If we look at the standard basis $\mathbb{E} = \{ 1, x, x^2 \}$, we observe that
$$p_\mathbb{E} = \begin{bmatrix} a \\ b \\ c \end{bmatrix},$$
so $\mathbb{E}$ is a basis of $P_2$ which satisfy your condition.
We can also show that this is the only basis which satisfy this condition: We notice that $$ p_\mathbb{B} = p_\mathbb{E} = T(\mathbb{B},\mathbb{E}) p_\mathbb{B} $$
where $T(\mathbb{B},\mathbb{E})$ denotes the basis change matrix from $\mathbb{B}$ to $\mathbb{E}$. But since $T(\mathbb{B},\mathbb{E})$ is invertible, we see that $T(\mathbb{B},\mathbb{E})$ must be the identity matrix which is equivalent to $\mathbb{B} = \mathbb{E}$.
