I have a polynomial equation:
$$y=(-5 \times 10^{-6} \times x^3)+(0.0004 \times x^2)+(0.0582 \times x)-0.4397$$
Is it possible to "rotate" this polynomial curve (maintaining the shape) around the point (10, 0.1773) so that it follows the gradient of the straight line:
$$y=(0.0222 \times x)+0.41797$$
Assume there is a new rotated coordinates $(x', y')$, and your polynomial curve has the same form $$y' = f(x') = -5\times 10^{-6}(x')^3 + 0.0004(x')^2 + 0.0582 x'-0.4397$$
Also, each point in $(x', y')$ is rotated from the original coordinates by an angle $\theta$ (to be obtained from your straight line):
$$\pmatrix{x'-10\\y'-0.1773} = \pmatrix{\cos\theta & -\sin\theta\\\sin\theta & \cos \theta} \pmatrix{x-10\\y-0.1773}$$
where $\tan \theta = 0.02222$.
Substitute these $x'$ and $y'$, in terms of $x$ and $y$, into the first equation.