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 $f(x)=x^2$ 
 A: 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.
A: 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
$$ \sin(25^\circ)x' + \cos(25^\circ)y' = (\cos(25^\circ)x' - \sin(25^\circ)y')^2 $$
A: To rotate the curve $y=f(x)$ around point $\langle a, b\rangle$ by angle $\theta$.
First express the curve as a parameterised vector equation.
$$\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix} t \\ f(t)\end{pmatrix}$$
Second, translate and rotate by matrix multiplication:
$$\begin{pmatrix}x_2 \\ y_2\end{pmatrix} =\begin{bmatrix}\cos \theta & -\sin \theta\\\sin\theta & \cos\theta\end{bmatrix}\times \begin{pmatrix} t-a \\ f(t)-b\end{pmatrix}+\begin{pmatrix} a \\ b\end{pmatrix}$$
