Finding basis for $P_2$ that satisfy conditions I want to find a basis $\mathbb{B}$ such that 
$[P_2]_\mathbb{B} =
    \begin{bmatrix} 
    p(0) \\ p(1) \\ p(2) 
\end {bmatrix}$
I assume general $P_2 = a+bx+cx^2$
Since $p(0)=a$
$p(1) = a+b+c$,   
$p(2)=a+2b+4c$
I want to find a basis $\mathbb{B}$ such that $[P_2]_\mathbb{B} =
    \begin{bmatrix} 
    a \\ a+b+c \\ a+2b+4c
\end {bmatrix}$
I tried $[{1}, {1+x+x^2},  1+2x+4x^2]$. I just tried using the general basis for $P_2$: $[1, x,  x^2]$ and logically deducing the basis that will satisfy the conditions in the task. Is the basis I've arrived at correct? 
 A: Starting out from the standard basis is good.
Calling $P_2$ the general element of $P_2$ and refer to the same element as $p$ in the next line is not good. We can calmly evaluate polynomials, i.e. if $p\in P_2$, then e.g. $p(1)$ makes perfect sense. So you just wanted to write $[p]_{\Bbb B}$.
Hint: We can get the coordinates of the standard basis vectors in the new basis. Write them up in the columns of a matrix and invert that matrix.
A: Goal: Find a basis $\mathbb{B}$ such that 
$$ [p]_\mathbb{B} = \begin{bmatrix} p(0) \\ p(1) \\ p(2) \end{bmatrix}$$
for every $p \in P_2$.
Let $p = a + bx + cx^2$ be an arbitrary element in $P_2$ with coefficients $a,b,c$. If we consider the standard basis $\mathbb{E} = \{ 1, x, x^2 \}$, we get
$$ [p]_\mathbb{E} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}.$$
On the one hand, as you showed us, we are looking for a basis $\mathbb{B}$ such that
$$ [p]_\mathbb{B} = \begin{bmatrix} a \\ a+b+c \\ a+2b+4c \end{bmatrix}$$.
But on the other hand, we know that
$$
T_{\mathbb{E},\mathbb{B}} [p]_\mathbb{E} = [p]_\mathbb{B}
$$
where $T_{\mathbb{E},\mathbb{B}}$ the basis change matrix from $\mathbb{E}$ to $\mathbb{B}$.
As it turns out, we must have $$T_{\mathbb{E},\mathbb{B}} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{bmatrix}$$
whose inverse is
$$
T_{\mathbb{B},\mathbb{E}} = \begin{bmatrix} 1 & 0 & 0 \\ -\frac{3}{2} & 2 & -\frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix}
$$
which is equivalent to $\mathbb{B}$ being the basis $\left[1 -\frac{3}{2}x + \frac{1}{2}x^2 , 2x-x^2, -\frac{1}{2}x+\frac{1}{2}x^2 \right]$.
