Identifying a polynomial by its value at specific points I am writing this question as a self-answer, because I suspect, although I
do not know for sure, what the OP (i.e. original poster) was trying to ask
in this question
which is now closed (make that now deleted).
I had intended to give a long-winded response comment, in an answer box, but the question was closed (and then deleted).
Further, the question that I think that the OP was trying to ask seems worthwhile.
So I will guess what the question was, present the question below, and then provide
an answer.  Note that I might be misinterpreting the OP's actual intent.
The Problem:
Find a polynomial $f(y)$ in the variable $y$ that satisfies the
following $3$ constraints:

*

*$f(1) = 12.$

*$f(2) = 6.$

*$f(3) = 18.$
Note: before I posted, I searched for duplicates, and found
this question.
However, that question didn't provide an explicit step-by-step solution.
Therefore, I feel that this question is not precisely a duplicate.
 A: You cannot find the polynomial that satisfies the given constraints, because if $f(x)$ is a solution, then, for any polynomial $g(x)$, the polynomial
$$
f(x)+(x-1)(x-2)(x-3)g(x)
$$
will satisfy the same constraints. But these are all the solutions! Indeed, if $f_1(x)$ is another solution, then the polynomial
$$
p(x)=f_1(x)-f(x)
$$
has the properties that $p(1)=p(2)=p(3)=0$ and so it is divisible by $x-1$, $x-2$ and $x-3$, hence by their product (use the remainder theorem) and therefore, writing $p(x)=(x-1)(x-2)(x-3)g(x)$, we get
$$
f_1(x)=f(x)+(x-1)(x-2)(x-3)g(x)
$$
as claimed.
Three conditions can determine at most three unknowns, so you can indeed look for solution of degree at most $2$. Using the system you do in your answer is good, but you can also directly use a Lagrange interpolating polynomial
\begin{align}
f(x)
&= 12\frac{(x-2)(x-3)}{(1-2)(1-3)}
 + 6\dfrac{(x-1)(x-3)}{(2-1)(2-3)}
 + 18\frac{(x-1)(x-2)}{(3-1)(3-2)}
\\[6px]
&=6(x^2-5x+6)-6(x^2-4x+3)+9(x^2-3x+2)
\\[12px]
&=9x^2-33x+36
\end{align}
Can you see why the polynomial satisfies the constraints? If you evaluate it at $1$, the second and third summand become zero, whereas the first summand is precisely $6$. Similarly when evaluating at $2$ and $3$.
A: 
Find a polynomial $f(y)$ in the variable $y$ that satisfies the
following $3$ constraints:



*

*$f(1) = 12.$

*$f(2) = 6.$

*$f(3) = 18.$


Edit
Per the answer of egreg, there is no unique solution.  However, as indicated below, if you arbitrarily decide that $f(y)$ must be a 2nd degree polynomial, then the answer is unique.

Just as two constraints similar to the above may be defined by a 1st degree
polynomial, three such constraints may be defined by a 2nd degree
polynomial.  In fact $n$ such constraints may be defined by a
polynomial of degree $(n-1)$.
This means that the simplest solution is to assume that $f(y)$ is a 2nd degree polynomial.  Under this assumption, real numbers $a,b,c$ must be chosen so that
$f(y) = ay^2 + by + c.$  This leads to the following three equations in
three unknowns:
$a + b + c = 12~:~~~~~$ based on $y = 1$.
$4a + 2b + c = 6~:~~~~~$ based on $y = 2$. 
$9a + 3b + c = 18~:~~~~~$ based on $y = 3$.
Subtracting the 1st equation from the 2nd gives 
$3a + b = -6$.
Subtracting the 1st equation from the 3rd gives 
$8a + 2b = 6 \implies 4a + b = 3.$
Subtracting the 1st derived equation from the 2nd derived equation gives 
$a = 9.$
Plugging this back into the 1st derived equation gives 
$b = -33.$
Plugging these $a,b$ values into the 1st original equation gives 
$c = 36.$
Therefore $f(y) = 9y^2 - 33y + 36.$
