Is any graph that is symmetric to the origin and to the y-axis also symmetric to the x-axis?

Question

Also, how would I support my answer graphically (what kind of graph would I draw) and how would I support the answer algebraically? Thank you.

Answer 1

YES. Graphs that are symmetric to origin and y-axis are symmetric to x-axis.

Consider a square with the diagonal along the y axis .
Let the meeting point of the two diagonals of the square( i.e the centre of the square ) be at the origin, then this will be symmetric to the origin, y-axis as well as the x-axis.

Answer 2

Symmetric about the y-axis implies an EVEN function, as in f(x) = f(-x). Meanwhile symmetric about the ORIGIN is the same as being an ODD function, f(x) = -f(-x).

A clear example occurs in both examples that has been thus far overlooked, the zero function f(x) = 0. It is the same when reflected about the y-axis, the x-axis, and the origin. Trivial but sufficient.

Is ANY graph though? No, not at all.

Answer 3

A graph is symmetric to the origin if you replace y with -y and x with -x and still have the same graph. A graph is symmetric with the y-axis if you replace x with -x and you still have the same graph. A graph is symmetric to the x-axis if you replace y with -y and still have the same graph.

If your graph is symmetric to the origin, then a point (x, y) on your graph means that the point (-x,-y) is also on the graph. Since the graph is also symmetric to the y-axis you can then replace x with -x to get the point ( -(-x), y) which is equivalent to (x,-y).
Therefore, (x,-y) is also on the graph.

To test for x-axis symmetry we replace y with -y in our original point (x,y) to get (x-,y). But we just showed above that the point (x,-y) is on the graph.

This shows that a graph with origin symmetry and y-axis symmetry must also have x-axis symmetry

Answer 4

One answer is Circle

I am sure you can proof it geometrically.

Algebraic proof:

1- A symmetric function (f(x,y)) to X axis:

f(x,y)=f(x,-y)

Circle:

f(x,y): x^2+y^2=C^2
f(x,-y): x^2+(-y)^2=x^2+y^2=C^2

2- A symetric function (f(x,y)) to Y axis:

f(x,y)=f(-x,y)

circle:

f(x,y): x^2+y^2=C^2
f(x,-y): (-x)^2+(y)^2=x^2+y^2=C^2

3- A symetric function (f(x,y)) to origin axis:

f(x,y)=f(-x,-y)

Circle:

f(x,y): x^2+y^2=C^2
f(-x,-y): (-x)^2+(-y)^2=x^2+y^2=C^2


Based on above method you can find other answers like:

f(x,y):x^4+y^4=C^4
A square or a rectangle also has such characteristics but they do not have continuous functions.

Answer 5

circle with centre at origin

Answer 6

yes,,,,an iso linearic sphere

Answer 7

no. that would be and is impossibe.