Q:

graphing $f(ax-b)$

I want to know which of the following methods is right for graphing $f(ax-b)$ from $f(x)$ and why:
Method 1. First Horizontally Translate it ($f(x)$) by $b$, then Horizontally Stretch/Compress it by $a$.
Method 2. First Horizontally Stretch/Compress it by $a$ then Horizontally Translate it by $b/a$.
Method 3. First Horizontally Translate it by $b/a$ then Horizontally Stretch/Compress it by $a$.

A:

Try turning $y=f(ax-b)$ inside out by applying the inverse of $f$ (i.e. $f^{-1}$).
$$y=f(ax-b)$$
$$f(ax-b)=y$$
$$f^{-1}(f(ax-b))=f^{-1}(y)$$
$$ax-b=f^{-1}(y)$$
$$ax=f^{-1}(y)+b$$
$$x=\frac{1}{a}(f^{-1}(y)+b)$$
$$x=\frac{1}{a}f^{-1}(y)+\frac{b}{a}$$
It's clear from this approach that we have a compression (with a factor of $a$) "along the x-axis". Notice that I didn't say "along the vertical or horizontal axis". That's because, at this point, I'm looking at the graph with my head tilted to the side (and I'm also looking at it in the mirror, so it gets very confusing). We also have a  positive shift of $\frac{b}{a}$ along the x-axis.
The one problem with this approach is that not all functions are 1-to-1 (and thus invertible). But we can get around this problem just by defining $f$ as a piece-wise function, and performing this process for each of the pieces. Note that each piece function will have their own inverse. You will find, however, that each 1-to-1 piece will have exactly the same transformations applied to them, namely a compression of $a$ and a positive shift of $b/a$ along the x-axis.

