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You know how we were told to use -b/2a to find the x-coordinate of the vertex for a quadratic?

Well, it turns out that you were really just solving for when the derivative equals zero. I don't know if anyone has noticed this but isn't that amzaing?

By Tongos (Tongos) on Tuesday, September 28, 2004 - 11:05 pm: Edit

yeah, i noticed this a long time ago. first, you find the vertex by doing the derivative. because parabolas are symmetric on each side, and you know how steep it is on each side, then you can find the zeros. because the vertex is like the starting point of the parabola. so treat it as if it was at the origin

By Tongos (Tongos) on Tuesday, September 28, 2004 - 11:07 pm: Edit

oh, that's something else, nevermind.

By Thermodude (Thermodude) on Tuesday, September 28, 2004 - 11:12 pm: Edit

yeah...that's pretty neat.

By Feuler (Feuler) on Tuesday, September 28, 2004 - 11:22 pm: Edit

That's what is so neat about math- it is all 100% consistent. You can go about a problem a bunch of completely different ways but you always get the exact same result (and by trying a new appraoch you usually learn something new about the old one).

By Welshie (Welshie) on Wednesday, September 29, 2004 - 01:11 pm: Edit

Or, you could do it a different way and get a completely different looking answer but have it end up equivalent to other correct answers.

Also, another epiphany to some, area and perimeter are directly connected (via integration/differentiation). You need some manipulation to see it but trust me, it's there.