Can a curve represented by a polynomial equation having n arbitrary constants, be uniquely determined by n-1 distinct points? Let there be a polynomial $f(X)$, $X \in R^m$  with $n$ unknown arbitrary constants. 
Consider , the equation  $ f(X) = 0$. Now, given $k$ distinct points in $R^m$   which satisfies the aforementioned equation such that $k \leq n$,  does it trivially follow or if not, how should it be proved(or provided with a counter example) that the $k$ linear equations we get form an independent system of equations?
 A: In one dimension, knowing $n$ different zeroes of a degree-$n$ polynomial will allow you to determine its $n+1$ coefficients up to a common constant factor. (Of course not all polynomials have $n$ different zeroes, but never mind that).
Things are not so nice in higher dimensions. For example if I tell you that $f(x,y)$ is a second-degree polynomial, and that
$$ f(0,0) = f(0,1) = f(0,2) = f(0,3) = f(0,4) = f(0,5) = 0$$
then you have just as many zeroes as there there are coefficients to find (a quadratic in 2 variables has 6 coefficients), but you still haven't learned anything useful about the relation between the coefficients of $x$, $xy$ and $x^2$.
A: This is coming from the determinant of the Vandermonde matrix.
A: The dimension of $d$-variate polynomial of degree $m$ is $Q=\dim \pi_m(\mathbb{R}^d)={m+d\choose d}$.
One would hope that $Q$ points -- roots would be enough to uniquely determine the polynomial. Indeed for $m<2$ this is the case. But for $m\geq2$ the layout of points matter, some layouts (e.g. "collinear") do not work, are not enough, the points must be more spread out, unisolvent (*). (It's a consequence of Mairhuber–Curtis theorem proving that there are no Haar spaces in $m\geq2$).
(*) Points are called $\pi_m(\mathbb{R}^d)$-unisolvent if the zero polynomial is the only polynomial from $\pi_m(\mathbb{R}^d)$ that vanishes on all of them.
