How to rotate any graph by any angle about any point? How can I do exactly as the title asks? I'm tearing my hair out trying to figure this out. I watched this video, and I now understand how to rotate about the origin, but I tried using the same steps to find the equation for any point, using h and k to represent the coordinates of the "new origin", and I got lost. I got to this point, and have no idea how to simplify, or if its even possible at all.

Any help at all will be greatly appreciated. Thank you.
To use this method one should convert $y=F(x)$ form into two separate parametric equations of form $x=f(t), y = g(t) $. For example 
$$ y^2 = 4 f x ; \rightarrow x= ft^2, y=2 ft $$

Next, a translation and a rotation  through angle $\theta$ can be done as you know$ (x,y)$ shifts $(h,k)$ between required points.

$$ x_1= x \cos \theta - y \sin \theta+ h\, ; y_1= x \sin \theta + y \cos \theta + k. $$

Next, you can see yourself if eliminating $\theta$,( which can be done) is desirable or worthwhile.
Do you have a function you're working with? i.e $f(x)$ = something?

If so, you can translate the point you're rotating about to the origin, rotate and then translate back. 

To translate vertically by $\pm k$, simply look at $f(x) \pm k$. To translate horizontally by $\pm h$, simply look at $f(x \mp h)$

As an example of translating, say we have the function $f(x) = (x-2)^2 + 4$ and we want to translate it so the point $(2, 4)$ coincides with the origin. Using the notation in the previous paragraph, $k = -4$ and $h = -2$. To take care of the vertical translation we look at $g(x) = f(x) - 4 = (x-2)^2$. To take care of the horizontal translation we look at $h(x) = g(x+2) = x^2$. Now our function has been translated to the origin. You can rotate $h(x)$ and then translate its rotated version vertically by $+4$ and horizontally by $+2$