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.