TITLE: Prove there exists a unique $n$-th degree polynomial that passes through $n+1$ points in the plane
QUESTION [21 upvotes]: I know given two points in the plane $(x_1,y_1)$ and $(x_2,y_2)$ there exists a unique 1st degree (linear) polynomial that passes through those points. We all learned in Algebra how to find the slope between those points and then calculate the y-intercept.
To take it down a notch, given the point $(a,b)$, the unique 0th degree polynomial that passes through it is $y=b$.
My conjecture is that given three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, there exists a 2nd degree (quadratic) polynomial that passes through these points, and furthermore, that polynomial is unique. I wonder, how would one determine the equation of this quadratic?
If my conjecture is correct, a corollary would be the generalization that given any $\left(n+1\right)$ points in the plane, there exists one unique $n$th degree polynomial that passes through those points.
Please prove, or disprove with a counter-example.
Further Readings:

658789 is a related question but I'm not sure if it's exactly what I'm looking for.
This very cool interactive web app lets you drag points around and shows the polynomial that goes through them

REPLY [3 votes]: johannesvalks' answer proves existence but not uniqueness.  The Lagrange interpolation polynomials used will have degree $\le n$, as each is the sum of polynomials. Given two polynomials $P,Q$ with $\deg(P),\deg(Q)\le n$ both passing through $n+1$ distinct points $(x_i,y_i)$ with $1\le i\le n+1$, divide to get $P=(x-x_i)P_i+y_i$ and $Q=(x-x_i)Q_i+y_i$, for some $P_i,Q_i$, so that $$P-Q=(x-x_i)(P_i-Q_i).$$  Then $P-Q$ has $n+1$ distinct roots, while $P$ and $Q$ have at most $n$, so $P-Q$ must be the zero polynomial.