Maximizing the area of a triangle with its vertices on a parabola. So, here's the question:
I have the parabola $y=x^2$. Take the points $A=(-1.5, 2.25)$ and $B=(3, 9)$, and connect them with a straight line. Now, I am trying find out how to take a third point on the parabola $C=(x,x^2)$, with $x\in[-1.5,3]$, in such a way that the area of the triangle $ABC$ is maximized.
I have pretty good evidence by trial and error that this point is $(.75, .5625)$ but I have no idea how to prove it. I was trying to work with a gradient, and then Heron's formula, but that was a nightmare to attempt to differentiate. I feel like this is a simple optimization problem but I have no clue how to solve it. Any help is appreciated!
Thanks.

Assuming $A=\left(-\frac{3}{2},\frac{9}{4}\right),B=(3,9),C=(x,x^2)$, the area of $ABC$ is maximized when the distance between $C$ and the line $AB$ is maximized, i.e. when the tangent to the parabola at $C$ has the same slope of the $AB$ line. Since the slope of the $AB$ line is $m=\frac{9-9/4}{3+3/2}=\frac{3}{2}$ and the slope of the tangent through $C$ is just $2x$, the area is maximized by taking:
$$ C=\left(\frac{3}{4},\frac{9}{16}\right) $$
and the area of $ABC$ can be computed through the shoelace formula:
$$ [ABC] = \frac{729}{64}. $$
This area is just three fourth of the area of the parabolic segment cut by $A$ and $B$, as already known to Archimedes. Here we have a picture:

Also notice that in a parabola the midpoints of parallel chords always lie on a line that is parallel to the axis of symmetry. That immediately gives that $C$ and the midpoint $M=\left(\frac{3}{4},\frac{45}{8}\right)$ of $AB$ have the same $x$-coordinate. Moreover, it gives that the area of $ABC$ is the length of $CM$ times the difference between the $x$-coordinate of $B$ and the $x$-coordinate of $C$, hence $\frac{729}{64}$ as already stated.
