Quadratic System of Equations

Question

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$$ a = \left(y_3 - y_1 - \frac{2(y_2 - y_1)}{(x_2 - x_1)} \right) / \left((x_3^2 - x_1^2) -2(x_2 - x_1)\right) $$

Answer 1

A simpler approach: from $a x_1^2 + b x_1 + c = y_1$ and $a x_2^2 + b x_2 + c = y_2$ you get $a (x_1^2 - x_2^2) + b (x_1 - x_2) = y_1 - y_2$, so (noting that $x_1^2 - x_2^2 = (x_1 - x_2)(x_1 + x_2)$), $b = \dfrac{y_1 - y_2}{x_1 - x_2} - a (x_1 + x_2)$.
Similarly $b = \dfrac{y_1 - y_3}{x_1 - x_3} - a (x_1 + x_3)$. Subtract these: $$0 = \dfrac{y_1 - y_2}{x_1 - x_2} - \dfrac{y_1 - y_3}{x_1 - x_3} - a(x_2 - x_3)$$ so $$a = \dfrac{\dfrac{y_1 - y_2}{x_1 - x_2} - \dfrac{y_1 - y_3}{x_1 - x_3}}{x_2 - x_3}$$

Answer 2

Hint: The calculations are too elaborate. Why do we have $c_1$, $c_2$, and $c_3$? There is one and only one constant term $c$. It can be easily eliminated, resulting in two linear equations involving the unknowns $a$ and $b$.

For the elimination process, solving and substituting works, but is not a good idea. Take the first two equations, subtract. Now $c$ is gone. Take the second equation and the third, subtract. Again, $c$ is gone. We now have two linear equations for $a$ and $b$. Eliminate $b$ by multiplying the first new equation through by $x_2-x_3$, and the second new equation by $x_1-x_2$, so that the coefficients of $b$ match. Subtract, and solve for $a$. Do not expand anything you do not need to.

For a fancier approach that generalizes nicely, look up the Vandermonde Determinant.

Edit: After the comments about the multiple $c_i$, these were removed. The substitution process used is not the optimal approach, but it works. Back substitution is now needed to find the other two coefficients.