0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.