# Symmetry of Polynomials

## Main Question or Discussion Point

Hi! Brief question:

I wonder which conditions should a polynomial function of odd degree fulfill in order to be symmetric to some point in the plane.
Are there such conditions?

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Gib Z
Homework Helper
Of course there are! It's merely horizontal and vertical translations of the ordinary condition for an odd function, which is symmetric about (0,0).

For symmetry around (0,a) a function must satisfy:

$$a - f(x) = f(-x) - a$$

For symmetry around (b,0), $$f(b+x) = - f(b-x)$$.

Perhaps you can combine these conditions?

Ok the combination is clear:
$$-f(b-x)+a=f(b+x)-a$$

Thank you! :-)

Last edited:
Gib Z
Homework Helper
you meant $$a + f(x) = - f(-x) - a$$, didn't you?
because it must be equivalent to $$a + f(x) + a - f(x) = 0$$

Ok the combination is clear:
$$-f(b-x)+a=f(b+x)-a$$

Thank you! :-)
Now that you bring it up, im not sure what I meant lol. Its something like that, though im sure its NOT equivalent to the condition a=0 lol.

Ok the combination is clear:
$$-f(b-x)+a=f(b+x)-a$$

Thank you! :-)

Oh sorry I've realised that everything is ok at the very moment you've sent your post...and deleted the thing!

Gib Z
Homework Helper
No problem =]