Proving That The Graph Is Symmetric About The Origin

I know that the graph of an equation in x and y is symmetric with respect to the origin if replacing x by -x and y by -y yeilds the same equation.

How can one prove that if the graph is symmetric with respect to it's x and y axis's,then it's also symmetric with respect to the origin ?

How can i generalise this?

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If you replace $x$ only with $-x$, then... If you replace $y$ only with $-y$, then... so if you replace both at the same time, then... – J. M. Nov 26 '11 at 10:15
Ha Ha Ha..I know what your saying.Can this be proved using some algebra is what i'am asking? – alok Nov 26 '11 at 10:34
...you have some graph $g(x,y)=0$ with the properties $g(-x,y)=g(x,y)$ and $g(x,-y)=g(x,y)$. If one negates both variables, then what? – J. M. Nov 26 '11 at 10:39
then it would be -g(x,y) ?? – alok Nov 26 '11 at 10:43
The solution set of $g(x,y)=0$ is symmetric with respect to $O$ if $g(x,y)=0$ implies $g(-x,-y)=0$. The identity $g(-x,-y)\equiv g(x,y)$ is a sufficient condition for this symmetry, but it is not necessary. Consider, e.g., the function $g(x,y):=x y(e^x+e^y)$. – Christian Blatter Nov 26 '11 at 12:20