Rotate the graph of a function?

How do I rotate a graph of a function around a point, and show it in the related equation?

An example could be $f(x)=\lvert x\rvert$ (absolute Value) and $f(x)=x^2$

In case you are literally after a rotation of the graph of a function as a set, this works:

The graph of the function $f(x)=x^2$ is given by: $$ \{(x,x^2):x\in\mathbb{R}\} $$

Rotation by a degree $\alpha$ is given by the matrix: $$ \left( \begin{array}{cc} \cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{array} \right) $$

Applying this matrix to the above set, you get $$ \{(x\cos\alpha-x^2\sin\alpha,x\sin\alpha+x^2\cos\alpha):x\in\mathbb{R}\} $$

But as mentioned already, this is not the graph of a function anymore.

If you don't care about whether the result is a graph of a function, then you can certainly rotate the set of points that satisfy $y=x^2$.

The general plan is to start by writing down the correspondence between new and old coordinates. If you're rotating by 25° around $(0,0)$, for example, it would be something like

$$ \begin{align} x &= \cos(25^\circ)x' - \sin(25^\circ)y' \\ y &= \sin(25^\circ)x' + \cos(25^\circ)y' \end{align} $$ were $x'$ and $y'$ are new coordinates. You can then insert into your original equation and get $$ \sin(25^\circ)x' + \cos(25^\circ)y' = (\cos(25^\circ)x' - \sin(25^\circ)y')^2 $$