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Students tend to mix up signs when shifting function graphs around: consider $y=x^2$. To shift it one unit upwards ("increasing $y$"), you write $y=x^2+1$, to shift it to the right ("increasing $x$"), you write $y=(x-1)^2$. On the one hand, it's $+$, on the other hand you have to use $-$ for somewhat "the same thing".

It recently came to my mind that the interpretation of the function graph as all points fulfilling the relation $ \mathbb R^2\supset R = \{(x, y) ~|~ y=x^2\}$ might be more helpful for this matter:

  • To shift the graph vertically, you use $R_\text{v} = \{(x, y) ~|~ y+a=x^2\}$.
  • To shift horizontally: $R_\text{h} = \{(x, y) ~|~ y=(x+a)^2\}$
  • Negative values for $a$ shift in increasing direction for both cases.

Is there any literature or experience on this approach?

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    $\begingroup$ Can’t answer you exact question, but this makes me think I shall try presenting $(y-1)=(x-1)^2$ to help students grasp why $f(x-1)$ shifts the graph right. $\endgroup$ – lukejanicke Mar 3 '18 at 13:57
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    $\begingroup$ I agree that this is an issue. My intuition is that we aren't really scaling or translating the graph of the function. We are actually messing with the axes. $y=(x-1)^2$ is the graph of $y=x^2$, where the axes have been moved to the left by a unit (thus translating the graph to the right). Similarly $y-1 = x^2$ is the same graph, with the axes shifted down a unit. Solving for $y$ gives $y=x^2 + 1$, which is the "usual" up-shifted graph. I have had little luck conveying this idea to students (i.e. transforming axes instead), but I think it is the right intuition. $\endgroup$ – Xander Henderson Mar 3 '18 at 14:24
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    $\begingroup$ I always explain that a shift like $x \mapsto x - 1$ makes the function wait a unit before doing things; what used to happen at $x = 0$ now happens at $x = 1$, hence the rightward shift (if you think of the input as time, things now happen $1$ unit of time later than they would; to the right). Similarly, a transformation like $x \mapsto Bx$ makes the function go through inputs $B$ times as fast, hence it'll only take $\frac{1}{B}$ times as long as it used to. $\endgroup$ – pjs36 Mar 3 '18 at 17:34
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    $\begingroup$ If you think about circles (and other functions),the equations use - not +. The center of the circle (x-a)^2+(y-b)^2=r^2 is at (a,b). I would use - in your template to be consistent. y-k = a(x-h)^2, for example, has its vertex at (h,k). $\endgroup$ – Sue VanHattum Mar 8 '18 at 0:25
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Imagine the simple equation of a straight line: $ y=2x$.
You want to move this up one point (in the $Y$-direction), and you get: $y=2x+1$.

If you want to move this up one point in the $X$-direction, then you first need to write this with $x$ at one side: $x=y/2$.
So you get : $x=y/2+1$.
Writing this back to the $y=f(x)$ notation becomes: $y=2(x-1)$.

So, the appearance of the minus is a simple consequence of the fact that we write our functions in the notation $y=f(x)$.

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Sorry, I don’t know of any literature but I trialed this approach with a class recently. I felt strongly that applying different approaches to transformations in the x and y directions was less than helpful when they would subsequently meet, for example, circles where the centre is other than the origin and which can be considered as transformations. It didn’t work for the majority; the idea of adding a constant to the right hand side to increment y by that amount was too intuitive for them to relinquish.

My next approach will be to keep the changes to the right side of the equals sign with translations of a in the x direction resulting in $f(x-a)$ and translations of b in the y direction resulting in $f(x)+b$. I will go beck to making the link and showing consistency in the two directions when I later introduce circles.

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  • $\begingroup$ This was my experience trying it with a group of college freshmen recently as well. However, it was indeed after I had taught them the intuitive way. $\endgroup$ – Opal E Apr 25 '18 at 22:05
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The only "literature" I can think of is how nearly every algebra textbook I've seen presents the "point-slope" form of a linear function: $$y - y_1 = m(x - x_1)$$

Here, the translation of axes is applied the same to both variables.

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According to my experience concrete examples like the following are best suited to support the understanding. Starting from the axes $y = 0$ and $x = 0$ I have always shown the following, one after another of course. enter image description here

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