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akshayb
 one year ago
Best ResponseYou've already chosen the best response.0Please tell the answer.

hewsmike
 one year ago
Best ResponseYou've already chosen the best response.1May I assume that you mean test some function for symmetry across the x and y axes and the origin ? :) Generally it means applying some transformation and seeing if there is no change ie. swap x for x and see if f(x) = f(x) so with \[f(x) = x^{2}\]then \[f(x)=(x)^{2} = (1)^{2}x^{2}= x^{2}=f(x)\]if we are plotting this function in two dimensions as \[y =x^{2}\] dw:1359934489024:dwthen this implies symmetry of reflection across the yaxis ( pardon my drawing skills ). Symmetry across the xaxis is likewise. As regards the origin : then that is a reflection across the yaxis and then another reflection across the xaxis. Thus a circle dw:1359935010242:dwis symmetric across x and y axes and hence the origin too. You can deduce this from the equation for a circle, say : \[x^{2}+ y^{2} = 1\]is invariant if we swap x for x and y for y \[(x)^{2} + (y)^{2} = (1)^{2}x^{2} + (1)^2y^{2} = x^{2} + y^{2}\]To be complete I ought mention that reflections across the x then y axes ( or vice versa ) is equivalent to a rotation of 180 degrees around the origin ....

hewsmike
 one year ago
Best ResponseYou've already chosen the best response.1Err, the circle was in that second graphic earlier today.
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