fitting a quadratic to 3 coordinates

Question

say I have 3 points on the plane (Cartesian coordinate system), (a,b), (c,d) and (e,f), I am fairly certain that there is one unique quadratic curve which passes through each point. what is the formula $(y=ux^2+vx+w)$ for the curve in terms of a, b, c, d, e and f?

Answer 1

If you know how to solve systems of equations thn you should be able to do this. You get for example $0 = Aa^2 + Ba + C - b$. Three equations in the unknowns $A,B,C$, so yes the solution is unique (assuming the 3 points are).

Answer 2

A generic quadratic curve in the plane has equation $$ ax^2+by^2+cxy+dx+ey+f=0 $$ So three points are not sufficient to fix all the six coefficients. If you want a particular quadratic as a parabola with symmetry axis parallel to $y$ axis, i.e with the equation $y=ux^2+vx+w$ then you have only tree coefficients and tree points suffice. To find the coefficients symply substitute the coordinates of the points in the equation so you have three linear equations with the coefficients as unknowns, solving the system you find the coefficients.