function inversion and the horizontal shift I am currently doing inverse functions and graphing radical equations of the form $y=a\sqrt{x-h}+k$ with my algebra class and one of my students asked me the following question.
"Why is it that we shift left when $\sqrt{x+h}$ and into the negative x-values but we shift up when $\sqrt{x}+k$ and into the positive y-values?"
I explained to them that part of it can be seen as a result of the inversion process of some functions/relations.  So I recalled the vertex form of a parabola, something they studied at the beginning of the year and I put it on the board in a specific function, as
$$y=x^2+3$$.
Then I graphed it and showed them that it was a parabola shifted up three units.  Then I inverted it by switching $y$ and $x$ and solving for $y$.  So
$$x=y^2+3 \Rightarrow x-3=y^2 \Rightarrow y=\pm\sqrt{x-3}$$
Now this is the inverse of our original function (although in its current form it is NOT a function) and notice since they are inverses they are symmetrical along the $y=x$ line.  Now look at the graphs and notice that even though our argument under the radical is $(x-3)$ we shift right into the positive reals.
I don't feel like this is reason enough to justify the why.  I'm also not sure that I satisfied the asker.  They saw both functions, they saw the symmetry on the $y=x$ line, so after learning about inverses, I think they sort of understood the relationship, but hopefullly someone can provide me with a bit of better insight as to why.  Please remember this question is ultimately for me and I do understand math well.  If I understand, then I can better teach.  
EDIT:  I'm specifically looking for teachers who have taught this material before who have maybe had to explain this.
 A: I explain the "horizontal shift" this way:  when we graph $ \ y = f(x+h) \ , $ we are composing $ \ f(x) \ $ on the function $ \ x + h \ . $  This is to say that we are first adding $ \ h \ $ to $ \ x \ , $ evaluating the function $ \ f \ $ at $ \ x + h \ , $ and then plotting  the result at $ \ x \ . $  This has the effect of reading off values of the function that lie to the right of $ \ x \ $ and moving them back to the left by $ \ h \ $ units.  Similarly, $ \ f(x - h ) \ $ moves the graph of $ \ f(x) \ $ to the right by $ \ h \ $ units, as we are reading off values of the function that lie to the left of $ \ x \ , $ in order to plot them at $ \ x \ . $
Horizontal shifts are definitely the more puzzling of the two.  I rarely see anyone have difficulties understanding what adding or subtracting a value directly to $ \ f(x) \ $ is going to do to the graph.  I also don't find anything "wrong" with your explanation in terms of inverse functions (a sort of "diagonal mirror" argument), but I think a student would typically need more experience with graphing functions (like so many things, a skill even university students don't have nearly enough practice with) to appreciate your description.
A: I like to think of it this way.  Imagine that you have drawn a function say $y=\sqrt{x}$ on a grid.  But then you decide to "magically" take your grid, including the origin and move it down by 2 units.  You are replacing every $y$-value with $y-2$  What will happen to the graph if you have "moved" your origin down 2 units. The graph will appear to be two units "higher"
The output $y$-values on the function will now be $y=f(x)+k$ where $k=2$
By replacing $y$-values with $y-2$, you are moving the plane down, and in effect moves the function "up"
When we notice it this way, what happens to $x$ and $y$ is really the same.
$f(x-h)$  is like moving the origin to the left $h$ units, which, in turn, moves the function right by $h$-units.
So I see the replacements acting the same way.
$y \to y-k$ is thinking $y=f(x)+k$ moving $k$ units up
$x \to x-h$ is moving $h$ units right.
In relation to your original function.
$y=\sqrt{x}+k$ comes from the parent function $y=\sqrt{x}$
but with a replacement of $y\to y-k$ This will result in the function appearing up.
$y=\sqrt{x+h}$ comes from the parent function $y=\sqrt{x}$ but is a replacement of $x \to x+h$ (moving the plane right) so results in the function appearing left.
In summary,
I think we are transforming the plane and the function coordinates respond to the change in the coordinate system and scale.
I wish I could find research to support this.
