Graphic

Yes I know the clumsy graphic I've chosen is nothing to do with completing the square as such...it is, of course differentiating the curve of y=f(x) (though it'll be a -x2 curve) from first principles, demonstrating how you find the gradient at A by steadily chopping smaller and smaller slices of the distance between A and B until the difference tends to 0 (ie at A). You've therefore used straight line gradients to find a pretty good approximation of the gradient at A (and therefore at any point on the curve).


Oooh, I think I am going to have to be careful here..I don't feel quite so....dogmatic as usual.

UPDATE: actually, looking at it, it seems to be a sort of cubic graph thingy in a positive x stylee. OR it might be something else. This doesn't really affect anything, you can still take dy/dx to find the gradient at any point.

I really do feel on eggshells here. Wow, this is weird.