# 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.

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.

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.