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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).

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1 Answer

Putting, $4a+b=1$ in $4a+b+b+c=3$ gives $b+c=2$, putting this in $a+b+c=3$ gives $a=1$

Now, $b=1-4a=-3$ and $c=2-b=5$

This method is primarily substitution method of solving such system of linear equations.It yields often quick solutions but might not be very easy to apply in all cases.

General method to solve such equation can be found by searching linear equation in $3$ variables if you are familiar with linear algebra (matrices, determinants, etc).

EDIT: Welcome to math.SE. :-)

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Thanks so much -- I can't believe I missed the obvious substitution -- breaking up the 2B into B + B! –  Jesse Nov 15 '12 at 14:15
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