Given f(x), find an expression to check whether f(x) has rotational symmetry about any arbitrary point (h, v).
If [tex]f(x) = f(-x)[/tex] then the function is symmetrical about the y-axis.
If [tex]f(x) = -f(-x)[/tex] 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 [tex]f(x) = x^3 + v[/tex] because I know it has this symmetry.
[tex]f(-x) = (-x)^3 + v[/tex]
[tex]f(-x) = -x^3 + v[/tex]
[tex]-f(-x) = x^3 - v[/tex]
Then I found the difference between f(x) and -f(-x).
[tex]f(x) - (-f(-x)) = (x^3 + v) - (x^3 - v)[/tex]
[tex]f(x) + f(-x) = 2v[/tex]
[tex]f(x) = -f(-x) + 2v[/tex]
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 [tex]f(x) = -f(-x+2h)[/tex], 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 [tex]f(x) = -f(-x+2h)+2v[/tex] for a real point (h, v), then the function's graph is point-rotational about (h, v). Is this correct?
p.s. I made this problem up for myself, if that's okay.