What is the Maximum Angle to Rotate a Parabola and Still Graph as a Function?

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Discussion Overview

The discussion centers on determining the maximum angle at which a parabola (specifically y=x²) can be rotated while still being graphed as a function, meaning it must have only one y-value for each x-input. The scope includes mathematical reasoning and conceptual clarification regarding the properties of functions and their graphical representations.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the maximum angle of rotation, suggesting it might be zero since any rotation could potentially violate the function's definition.
  • Another participant proposes that a slight rotation might be possible but acknowledges uncertainty about where the cutoff point lies.
  • A different participant argues that any rotation results in a quadratic equation that yields two y-values for each x, thus confirming that the parabola cannot be rotated without losing its function property.
  • One participant reflects on their intuitive exploration of the problem, initially believing a small rotation might work, but ultimately agrees with the mathematical reasoning presented by another participant.
  • Another participant expresses agreement with the logic of the previous arguments, indicating a feeling that the answer might indeed be zero, but they remain unsure.

Areas of Agreement / Disagreement

Participants express differing views on the maximum angle of rotation. Some argue that any rotation makes the parabola no longer a function, while others speculate about the possibility of a slight rotation. The discussion remains unresolved regarding the exact cutoff point for rotation.

Contextual Notes

The discussion involves assumptions about the nature of functions and the implications of rotating a parabola. There are unresolved mathematical steps regarding the exact conditions under which the parabola retains its functional properties.

Vorde
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What is the maximum angle (degrees or radians) that you can rotate the basic parabola (y=x2) so that it can still be graphed as a function (y=...) with only one possible y-value per x-input.
 
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I think it's 0, because when you include the xy factor, it doesn't become a function anymore.
 
But more abstractly, I think it's possible to do a slight rotation, but there's an obvious cutoff point. I'm curious where that cutoff point is, it could be zero, I can't quite picture it well enough.
 
No, that's not correct. Any rotation at all makes it no longer a function.

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 two values of y for each x.
 
Vorde, I could not fault the logic presented by Hallsofivy, but it didn't FEEL right, so I played w/ it a bit from what I thought of as a more intuitive way of looking at it thinking it would show that at least a small rotation would work, but it clearly doesn't.

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.
 
Both what HallsofIvy and phinds said make perfect sense to me. I had a feeling the answer might be zero, but I couldn't convince myself either way, thanks to both of you.
 

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