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 $\frac{\mathrm dy}{\mathrm dx} = 2AX + B$, 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).