# Proving symmetry algebraically about origin and axes.

• Feb 18th 2011, 05:20 PM
jonnygill
Proving symmetry algebraically about origin and axes.
Hello!

how could i show that any graph that is symmetrical to the origin as well as the y-axis must also be symmetrical to the x-axis?

i understand how to determine if a given function is symmetrical to the origin, x-axis, and y-axis

intuitively i believe that the statement in the above post is true. but i have no idea how to prove this.

edit: ok, so i've thought about it a bit. i'm thinking you would first pick a point on the graph, lets say in QI. say it is (2,1). then rotate it 180 degrees about the origin giving a point of (-2,-1). starting with that same point again, (2,1) i would know that another point exists on the graph that is (-2,1) or the y-axis symmetrical point. i see that the points created with the point symmetry of the origin and the line symmetry of the y-axis are symmetrical across the x-axis.

however, i could draw a graph that goes through each of these three points and does not go through (2,-1). a point that is important to show x-axis symmetry.

could someone please explain to me how to prove this algebraically? or prove that it is not true algebraically?

thank you!
• Feb 19th 2011, 05:19 AM
HallsofIvy
A graph is "symmetric about the orgin" if and only if "whenever (x, y) is on the graph, so is (-x, -y)".

A graph is "symmetric about the y axis" if and only if "whenever (x, y) is on the graph, so is (-x, y)".

If a graph is symmetric about the origin and symmetric about the y axis, then, if (x, y) is on the graph so is (-x, -y) because of the symmetry about the origin, and then so is (x, -y) because of the symmetry about the y-axis.

But that says "whenever (x, y) is on the graph, so is (x, -y)" which is the same as "symmetric about the x axis".