Is it possible to find function that contains every given point? Let say we have a arbitrary number of given points and there is at least one function, for which every point lies on its graph. Is it possible to find that function using only X and Y coordinates of every given point?
Example:
We are given points $A(0,1)$; $B(0.27;0)$ and $C (3.73;0)$. All of these points lie on the graph of the function $f(x) = x^2 - 4x + 1$. Is it possible to find this function using only X and Y coordinates of the point A,B and C.
Note: The number of point is can be really big and also the exponent of the function.
 A: There are many such functions so in general it's not possible to recover the one you started with. 
For example if you had a set of points $\{(n,n): n\in\mathbb N\}$ both the functions $f(x) = x$ and $g(x) = x+sin(\pi x)$ would pass through all the points.
If you have a finite number of points you can find a polynomial that passes through them all. 
One way to find it would be the following algorithm. 
Let the coordinates of the points be $(x_1,y_1), (x_2,y_2),\dots, (x_N,y_N)$
First we find a function $f_1$ that passes through $(x_1,y_1)$ that's easy. $F_1(x) = y_1$.
Now suppose we've constructed $F_n$ that passes through $(x_1,y_1), \dots,(x_n, y_n)$ (so far we've only done this for $n=1$).
Let $$g_n(x) = \prod_{i=1}^n (x-x_i)$$
then $g_n(x_i) = 0$ for every $i\leq n$. Therefore for every value of $\lambda$ the function
$F_n(x_i) + \lambda g_n(x_i) = y_i$ for every $i\leq n$. 
So set $$\lambda_n = \frac{y_{n+1} - F_n(x_{n+1})}{g_n(x_{n+1})}$$ and $$F_{n+1}(x) = F_n(x) +  \lambda_n g_n(x).$$
Then $F_{n+1}$ passes through $(x_i,y_i)$ for every $i\leq n+1$.
So apply this step repeatedly until you get $F_N$, which passes through every point.
Let's try it with your equation I'm going to choose slightly different points so that we can use whole numbers. Set
$$\begin{array}{rlrl}
x_1 &=0 & y_1 &=1 \\
x_2 &=2 & y_2 &=-3 \\
x_3 &=5 & y_3 &=6 
\end{array}$$
It's easy to check that $x^2-4x+1$ passes through all three points. So let's follow the recipe.
First we set $F_1(x) = y_1 = 1$ and $g_1(x) = (x-x_0) = x$.
Then we can calculate $$\begin{array}{rl}\lambda_1 &= \frac{y_2 - F_1(x_2)}{g_1(x_2)} \\&= -2\end{array}$$
Hence $$\begin{array}{rl}F_2(x) &= F_1(x) + \lambda_1 g_1(x) \\ &= 1 -2x \end{array}$$
We can check that $F_2(0)=1$ and $F_2(2)=-3$.
Next we have $g_2(x) = x(x-2) = x^2-2x$
and we can work out $$\begin{array}{rl}\lambda_2 &= \frac{y_3 - F_2(x_3)}{g_2(x_3)} \\ &= \frac{6 - (-3)}{40} \\&= 1\end{array}$$
So $\lambda_2 = -\frac 3{15}$ and we have
Then we have $$\begin{array}{rl}F_3(x) &= F_2(x) + \lambda_2 g_2(x) \\ 
&= (1-2x) + (x^2 - 2x) \\
&= x^2-4x+1 
\end{array}$$ 
Then as we've run out of points we set $f(x) = F_3(x)$ and we are done.
A: Suppose you have $n$ given points. Assume a polynomial function of order $n-1$ with unknown constant coefficients. You'll now have to solve simultaneously $n$ number of equations for coefficients that make the polynomial curve pass through the points.
For example in your case, there are $3$ given points $A(0,1)$, $B(0.27,0)$ and  $C(3.73,0)$.
Now assume a polynomial of order $2$ with unknown constant coefficients.
$$y=A+Bx+Cx^2$$
Plug in the $3$ points to get $3$ equations in $A$, $B$ and $C$ and simultaneously solve them.
And indeed  $A=1$, $B=-4$ and $C=1$.
This can be employed for any arbitrary set of points, though the process gets increasingly tedious with large number of points. 
