It does! And you can even make the stronger statement that the resulting polynomial is degree at most $n$ (just like with quadratics).
If you've seen systems of linear equations, there's an easy way to see why this is the case. Suppose you want a polynomial $f(x) = a_0 + a_1 x + \ldots + a_n x^n$ such that $f(x_0) = y_0$, $f(x_1) = y_1$, \ldots, $f(x_n) = y_n$ where we assume $x_i \neq x_j$ if $i \neq j$. The equations $f(x_i) = y_i$ are really $n+1$ equations in $n+1$ unknown variables $a_0,\ldots,a_n$ (i.e., in the coefficients of the unknown polynomial $f(x)$):
\begin{align}
a_0 + x_0 a_1 + \ldots + x_0^n a_n &= y_0 \\
&\vdots\\
a_0 + x_n a_1 + \ldots + x_n^n a_n &= y_n
\end{align}
or in matrix form
$$
\begin{pmatrix}
1 & x_0 & \cdots & x_0^n\\
1 & x_1 & \cdots & x_1^n\\
\vdots&&&\vdots\\
1 & x_n & \cdots & x_n^n
\end{pmatrix}
\begin{pmatrix}
a_0\\a_1\\\vdots\\a_n
\end{pmatrix}
= \begin{pmatrix}
y_0\\y_1\\\vdots\\y_n\end{pmatrix}.
$$
The $(n+1)\times(n+1)$ matrix on the left is called the Vandermonde matrix of $x_0,\ldots,x_n$, and (with some work) you can show that its determinant is always equal to
$$
\prod_{0 \leq i < j \leq n}(x_j - x_i),
$$
which is nonzero if $x_j \neq x_i$ for $j \neq i$ like we assumed for obvious reasons above. It follows that the matrix is invertible, and so there is a solution to this system of equations:
$$
\begin{pmatrix}
a_0\\a_1\\\vdots\\a_n
\end{pmatrix}
= \begin{pmatrix}
1 & x_0 & \cdots & x_0^n\\
1 & x_1 & \cdots & x_1^n\\
\vdots&&&\vdots\\
1 & x_n & \cdots & x_n^n
\end{pmatrix}^{-1}\begin{pmatrix}
y_0\\y_1\\\vdots\\y_n\end{pmatrix}.
$$
What's more, notice that we've actually proven something stronger than what we set out to. Not only is there such a polynomial of degree $n$ or less, the polynomial is unique! That is,
For any set of $n+1$ points $(x_i,y_i)$ with $x_i \neq x_j$ for $i \neq j$, there is a unique polynomial $f$ of degree at most $n$ such that $f(x_i) = y_i$ for all $i$.
If you're interested, the Wikipedia page on polynomial interpolation algorithms provides some greater technical detail about the what these coefficients look like when written out explicitly and also how efficiently some related algorithms can be implemented. This includes so-called Lagrange interpolation method, which is derived from the above solution.
Assume you have given $n+1$ points $(x_1,y_1),\cdots, (x_{n+1},y_{n+1}).$ (Of course, $x_i\ne x_j$ if $i\ne j.$) A polynomial of degree $n$ is of the form $p_n(x)=a_nx^n+\cdots+a_1x+a_0.$ To study the existence and uniqueness of such a polynomial consider the system of linear equations:
$$
\left\{\begin{array}{ccc}
a_nx_1^n+a_{n-1}x_1^{n-1}\cdots+a_1x_1+a_0 & =& y_1\\ \vdots & &\\
a_nx_{n+1}^n+a_{n-1}x_{n+1}^{n-1}\cdots+a_1x_{n+1}+a_0 & =& y_{n+1}
\end{array}\right.
$$
We write the system as
$$
\begin{pmatrix}x_1^n & x_1^{n-1} &\cdots & x_1 & 1 \\ \vdots & \vdots & \ddots & \vdots \\ x_{n+1}^n & x_{n+1}^{n-1}& \cdots & x_{n+1} & 1\end{pmatrix} \begin{pmatrix} a_n \\ \vdots \\ a_0 \end{pmatrix}=\begin{pmatrix} y_1 \\ \vdots \\ y_{n+1} \end{pmatrix}
$$
Since the matrix of coefficients of the system is non singular (it is a Vandermonde matrix (see Vandermonde)) the system has a unique solution, that is, there exists one polynomial of degree $n$ through the $n+1$ given points, and it is unique.
Best Answer
Let us suppose that you have this general problem when you need the quadratic function which goes through three points $[x_i,y_i]$ and let, as written by dinosaur, $$y=f(x)=ax^2+bx+c$$ So, the three equations are $$y_1 = a {x_1}^2+b {x_1}+c$$ $$y_2 = a {x_2}^2+b {x_2}+c$$ $$y_3 = a {x_3}^2+b {x_3}+c$$ Subtracting the first to the second and the second from the third already eliminates $c$ and your are left with two linear equations for two unknowns. You could even eliminate from the first difference $b$ and plug it in the second difference and solve a linear equation in $a$. When $a$ is obtained, go backwards for getting $b$ and then $c$.
If you do the above, you will end with $$a=\frac{{x_1} ({y_3}-{y_2})+{x_2} ({y_1}-{y_3})+{x_3} ({y_2}-{y_1})}{({x_1}-{x_2}) ({x_1}-{x_3}) ({x_2}-{x_3})}$$ $$b=\frac{y_2-y_1}{x_2-x_1}-a (x_1+x_2)$$ $$c = y_1-a {x_1}^2-b {x_1}$$ Simple, isn't ?