I am given a quadratic equation: $$ y = Ax^2 + Bx + C $$ that passes through $(1,3)$ and $(2,3)$, and a tangent to the curve is $x - y + 1 = 0$ at $(2.3)$.
How do I find $A$, $B$, and $C$?
The derivative of $\mathrm dy/\mathrm dx = 2AX + B$, so at $x=2$, the slope of the tangent is $4A + B$, and from the givens we know that $4A + B = 1$. We also know that $$ 3 = A + B + C,\qquad 3 = 4A + 2B + C. $$
From there, how does one find $A$, $B$, and $C$?
(I can't seem to get the answers that make any sense from here).