I'm trying to define a quadratic that can pass through any 3 points. I've obviously done something wrong but can't figure out where. Any help would be appreciated.
$$ ax_1^2 + bx_1 + c = y_1 $$ $$ ax_2^2 + bx_2 + c = y_2 $$ $$ ax_3^2 + bx_3 + c = y_3 $$
Solve for C using the first equation
1$$ ax_1^2 + bx_1 + c = y_1 $$ 2$$ ax_1^2 + bx_1 - y_1 = -c $$ 3$$ -ax_1^2 - bx_1 + y_1 = c $$ 4$$ c = -ax_1^2 - bx_1 + y_1 $$
Now substitute C and solve be B using the second equation
5$$ ax_2^2 + bx_2 + c = y_2 $$ 6$$ ax_2^2 + bx_2 - ax_1^2 - bx_1 + y_1 = y_2 $$ 7$$ a(x_2^2 - x_1^2) + b(x_2 - x_1) = y_2 - y_1 $$ 8$$ b(x_2 - x_1) = y_2 - y_1 - a(x_2^2 - x_1^2) $$ 9$$ b = \frac{y_2 - y_1 - a(x_2^2 - x_1^2)}{(x_2 - x_1)} $$
Ok, now substitute B and C and solve A using the third equation
10$$ ax_3^2 + bx_3 + c = y_3 $$ $$ ax_3^2 + bx_3 - ax_1^2 - bx_1 + y_1 = y_3 $$
11$$ ax_3^2 + x_3\left(\frac{y_2 - y_1 - a(x_2^2 - x_1^2)}{(x_2 - x_1)}\right) - ax_1^2 - x_1\left(\frac{y_2 - y_1 - a(x_2^2 - x_1^2)}{(x_2 - x_1)}\right) + y_1 = y_3 $$
12$$ ax_3^2 - ax_1^2 + \frac{x_3(y_2 - y_1) - x_3a(x_2^2 - x_1^2)}{x_3(x_2 - x_1)} + \frac{-x_1(y_2 - y_1) + x_1a(x_2^2 - x_1^2)}{-x_1(x_2 - x_1)} = y_3 - y_1 $$
13$$ ax_3^2 - ax_1^2 + \frac{-x_1x_3(y_2 - y_1) + x_1x_3a(x_2^2 - x_1^2)}{-x_1x_3(x_2 - x_1)} + \frac{-x_1x_3(y_2 - y_1)+ x_1x_3a(x_2^2 - x_1^2)}{-x_1x_3(x_2 - x_1)} = y_3 - y_1 $$
14$$ ax_3^2 - ax_1^2 + \frac{-x_1x_3(y_2 - y_1) + x_1x_3a(x_2^2 - x_1^2) - x_1x_3(y_2 - y_1) + x_1x_3a(x_2^2 - x_1^2)}{-x_1x_3(x_2 - x_1)} = y_3 - y_1 $$
15$$ ax_3^2 - ax_1^2 + \frac{(y_2 - y_1) - a(x_2^2 - x_1^2) + (y_2 - y_1) - a(x_2^2 - x_1^2)}{(x_2 - x_1)} = y_3 - y_1 $$
16$$ a(x_3^2 - x_1^2) + \frac{2(y_2 - y_1) - 2a(x_2^2 - x_1^2)}{(x_2 - x_1)} = y_3 - y_1 $$
17$$ a(x_3^2 - x_1^2) + \frac{2(y_2 - y_1)}{(x_2 - x_1)} + \frac{ -2a(x_2^2 - x_1^2) }{(x_2 - x_1)} = y_3 - y_1 $$
18$$ a(x_3^2 - x_1^2) + \frac{2(y_2 - y_1)}{(x_2 - x_1)} -2a(x_2 - x_1) = y_3 - y_1 $$
19$$ a(x_3^2 - x_1^2) -2a(x_2 - x_1) = y_3 - y_1 - \frac{2(y_2 - y_1)}{(x_2 - x_1)} $$
20$$ a((x_3^2 - x_1^2) -2(x_2 - x_1)) = y_3 - y_1 - \frac{2(y_2 - y_1)}{(x_2 - x_1)} $$
21$$ a = \left(y_3 - y_1 - \frac{2(y_2 - y_1)}{(x_2 - x_1)} \right) / \left((x_3^2 - x_1^2) -2(x_2 - x_1)\right) $$