How to tell if a function has rotational symmetry?

Question

How to spot rotational symmetry on(or in, of?) a function? If I have the function $f(x)={{5a^2+6ax+9x^2}\over {a+3x}}$, how can I know it has rotational symmetry about the point $(-{a\over 3},0)$?

Is there a general method to detect rotational symmetry?

Answer 1

Simplify it to $$ f(x)={{5a^2+6ax+9x^2}\over {a+3x}}=\frac{4a^2}{a+3x} + {{a^2+6ax+9x^2}\over {a+3x}}=\frac{4a^2}{a+3x} + {{(a+3x)^2}\over {a+3x}}\\ =\frac{4a^2}{a+3x} + (a+3x) $$ and substitute $y=a+3x$ then $$ f(y)=\frac{4a^2}{y}+y $$ which has the following property: $$ f(-y)=-\frac{4a^2}{y} -y = -f(y) $$ which means that $f(y)$ is an odd function, point symmetric around the origin, hence $f(x)$ is point symmetric around $0=a+3x$ so your symmetric center is shift accordingly...

Answer 2

$$f(-\frac{2a}3-x)={{5a^2+6a\left(-\frac{2a}3-x\right)+9\left(-\frac{2a}3-x\right)^2}\over {a+3\left(-\frac{2a}3-x\right)}}={{5a^2+6ax+9x^2}\over {-a-x}}=-f(x).$$