Explaining Horizontal Shifting and Scaling I always find myself wanting for a clear explanation (to a college algebra student) for the fact that horizontal transformations of graphs work in the opposite way that one might expect.
For example, $f(x+1)$ is a horizontal shift to the left (a shift toward the negative side of the $x$-axis), whereas a cursory glance would cause one to suspect that adding a positive amount should shift in the positive direction. Similarly, $f(2x)$ causes the graph to shrink horizontally, not expand.
I generally explain this by saying $x$ is getting a "head start". For example, suppose $f(x)$ has a root at $x = 5$. The graph of $f(x+1)$ is getting a unit for free, and so we only need $x = 4$ to get the same output before as before (i.e. a root). Thus, the root that used to be at $x=5$ is now at $x=4$, which is a shift to the left.
My explanation seems to help some students and mystify others. I was hoping someone else in the community had an enlightening way to explain these phenomena. Again, I emphasize that the purpose is to strengthen the student's intuition; a rigorous algebraic approach is not what I'm looking for.

The map $f$ already assigns $\color{Purple}{\sigma x\mapsto f(\sigma x)}$. In order to construct the assignment $\color{DarkBlue}x\mapsto \color{DarkOrange}{f(\sigma x)}$, we must apply the inverse $\sigma^{-1}$ in order to put $f(\sigma x)$ (originally located at $\sigma x$) "back" to $x$, that is
$$(\sigma^{-1},\,\mathrm{Id}):\big(\sigma x,f(\sigma x)\big)\mapsto \big(x,f(\sigma x)\big).$$
The above is, of course, too algebraic. So here's a colorful depiction of what's going on:
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As we can see, "putting it back over the $x$-value" does the inverse of the transform to the actual graph (squiggly line) of $f$.
