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^{2}) so that it can still be graphed as a function (y=...) with only one possible y-value per x-input.

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- Thread starter Vorde
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- #2

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I think it's 0, because when you include the xy factor, it doesn't become a function anymore.

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HallsofIvy

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Start with [itex]y= x^2[/itex]. With a rotation through an angle [itex]\theta[/itex] we can write [itex]x= x' cos(\theta)+ y' sin(\theta)[/itex], [itex]y= x' sin(\theta)- y' cos(\theta)[/itex] where x' and y' are the new, tilted coordinates.

In this new coordinate system, the parabola becomes [tex]x'sin(\theta)- y'cos(\theta)= (x'cos(\theta)+ y'sin(\theta))^2= x'^2 cos^2(\theta)+ 2x'y'sin(\theta)cos(\theta)+ y'^2 sin^2(\theta)[/tex].

Now, if we were to fix x' and try to solve for y' we would get, for any non-zero [itex]\theta[/itex], a quadratic equation which would have

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Here's how I got there. Think of a line that goes through the origin but really hugs the y axis. Let's say it has a slope of 1,000, and it has a sister line just on the other side of the y axis with a slope of -1,000. If neither of them hit the parabola, then clearly you could rotate it by that much. It's trivially easy to show though that they both DO hit the parabola (at x = 1,000 and x=-1,000 assuming the given example of y = x^2) so Hallofivy obviously had it right and that was all a waste of time mathematically, but it DID help me see more graphically why he is right.

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