Multiplication using Parabola

Theorem
Given the function $f\left(x\right) = x^{2}$ and two points $\left(A,B\right)\in \{\left(x,f\left(x\right)\right):x \in \Bbb{R}\}$ under the condition that $A_{x} \leq B_{x}\wedge (A_{x} \neq 0 \vee B_{x} \neq 0)$.

Then the line segment $AB$ will have a y-intercept of $-A_{x}B_{x}$.

Where $A_{x} \: \& \: B_{x}$ are the x-coordinates of points $A \: \& \: B$.

Proof
By subbing in the value $A_{x}$ into $f(x)$ the $y$-coordinate of $A$ is obtained.


 * $f(A_{x}) = A_{x}^{2}$

the same logic follows for the point $B$.

Since the line connecting $A \& B$ will be linear the slope $m$ can be defined:


 * $m = \dfrac{B_{x}^{2}-A_{x}^{2}}{B_{x}-A_{x}}$

The numerator is a Difference of Two Squares and so the whole fraction can be simplified:


 * $m = B_{x} + A_{x}$

The Result follows by choosing either $A$ or $B$ to sub the respective point's coordinates into the linear function with the slope of the line segment joining $A \: \& \: B$.

Solving for the y-intercept $b$ of $A_{x}^{2} = \left(A_{x}+b_{x}\right)A_{x}+b$ yields the result.

Proceeding with the point $A$.