How Can You Determine Point-Rotational Symmetry of a General Function?

[Unit]

Homework Statement

Given f(x), find an expression to check whether f(x) has rotational symmetry about any arbitrary point (h, v).

Homework Equations

If $$f(x) = f(-x)$$ then the function is symmetrical about the y-axis.

If $$f(x) = -f(-x)$$ then the function is point-rotational about the origin.

The Attempt at a Solution

I don't know how to deal with "general" functions, so I did this:

Let $$f(x) = x^3 + v$$ because I know it has this symmetry.
$$f(-x) = (-x)^3 + v$$
$$f(-x) = -x^3 + v$$
$$-f(-x) = x^3 - v$$

Then I found the difference between f(x) and -f(-x).

$$f(x) - (-f(-x)) = (x^3 + v) - (x^3 - v)$$
$$f(x) + f(-x) = 2v$$
$$f(x) = -f(-x) + 2v$$

This gives me point-rotation about (0, v). But how do I do this for just a "general" function?

Also, I can guess that point-rotation about (h, 0) will be something like $$f(x) = -f(-x+2h)$$, based on knowledge of transformations, but how do I show this as elegantly as above with (0, v)?

From this, I can guess that if $$f(x) = -f(-x+2h)+2v$$ for a real point (h, v), then the function's graph is point-rotational about (h, v). Is this correct?

Cheers,
Unit

p.s. I made this problem up for myself, if that's okay.

 

[HallsofIvy]

You don't mean "general function" you mean "general point".

Given a point (u, v) in the xy-plane, you can "translate" (u, v) to (0, 0) by subtracting u from x and v from y. f(x) becomes f(x- u) and y= f(x) becomes y- v= f(x- u) or, finally, y= f(x- u)+ v. Now, what happens if you replace x with -x?