1
$\begingroup$

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?

  • 0
    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...2011-11-26
  • 0
    Ha Ha Ha..I know what your saying.Can this be proved using some algebra is what i'am asking?2011-11-26
  • 0
    ...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?2011-11-26
  • 0
    then it would be -g(x,y) ??2011-11-26
  • 0
    How did that sign move to the outside of $g$? You know that $g(-x, \square) = g(x, \square)$ for any values of $\square$. Plug in $\square = -y$ and you get?2011-11-26
  • 0
    Now i'am confused2011-11-26
  • 1
    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 $g(x,y):=x y(e^x+e^y)$.2011-11-26
  • 0
    Ok..fair enough.2011-11-26

0 Answers 0