Symmetry in graphs

  • #1
For a graph of any function, one of following conditions is said to exist so as for it to be symmetric:
a graph is symmetric about y-axis if along with a point (x,y) a point (-x, y) exists.
a graph is symmetric about x-axis if along with a point (x,y) a point (x, -y) exists.
a graph is symmetric about origin if along with a point (x,y) a point (-x, -y) exists.
isn't it possible that one of the above condition is satisfied but still the graph is not symmetric. what i want to say is that can't it be possible for a curve to have such shape that it does not look symmetric but still passes through the two points highlighted by one of the conditions mentioned above?
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi Danish_Khatri! Welcome to PF! :wink:
For a graph of any function, one of following conditions is said to exist so as for it to be symmetric:
a graph is symmetric about y-axis if along with a point (x,y) a point (-x, y) exists.
a graph is symmetric about x-axis if along with a point (x,y) a point (x, -y) exists.
a graph is symmetric about origin if along with a point (x,y) a point (-x, -y) exists.
isn't it possible that one of the above condition is satisfied but still the graph is not symmetric. what i want to say is that can't it be possible for a curve to have such shape that it does not look symmetric but still passes through the two points highlighted by one of the conditions mentioned above?

I think it means for every (x,y) on the curve … :smile:
 
  • #3
Thanks for your help dear.... :-)
 

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