I'll just put my general solution I used on the test.

First, I specified a variable, z, to take the place of x to avoid confusion when comparing the curve to axes. Since the secant goes through (0, 0) and "another point on the curve", I used the point (z, (az^2+bz)). And as Titus said, b has to be zero, or the curve's extreme point won't go through the origin.

Then, I simply used the dy/dx definition to find the slope between the two points:

((az^2+bz)-0) / (z-0) = (az^2+bz) / z = (z(az+b))/z = az+b

To compare the secant to the tangent, I just derived the original curve to find the slope of the tangent at any given point for the same variables:

y=ax^2+bx
y'=2ax+b

Since the x value of the tangent is supposed to be the arithmetic mean value, (z+0)/2 = z/2

So just cram the x co-ordinate into the derivate function:

y'(z/2) = 2a*(z/2)+b = az+b

And there's my "proof" for why the slopes should always be the same for the secant and tangent, if real numbers are being used.

"Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram