# *Seeing* why horizontal shifts are reversed?

I know it's true, and understand why, but I can't see it. I feel stupid.

Perhaps I can't see it becaus too complex to hold at once, the two functions, the translation, the axes. Maybe it's not important to "see" it anyway, and I'm expecting too much... and knowing the rule and why is enough?

What If we start with function $f(x)$, then shift it vertically by $\Delta y$ and horizonally by $\Delta x$, to get another function $g(x)$, then

$$g(x) = f(x - \Delta x) + \Delta y$$

Although we add $\Delta y$, we subtract $\Delta x$ - this subtraction is the "reverse" of the question.

Why One way to think about it is that $g$ is sampling $f$ at another input point. To get to that other point, from the point of view of $g$, we go backwards, the opposite direction. When we describe $f$ being shifted to become $g$, it is from the point of view of $f$. So, this change in point of view is why we reverse the horizontal shift. If instead, we described the shift as where we came from, it would already be "reversed".

This creates another puzzle: why isn't $\Delta y$ reversed too? Because it's a translation of the output, after the change in point of view has already occurred.

In another way, the difference between horizonal and vertical translation is an artefact of notation. They are both reversed (or, from the point of view of the new function), if notated as:

$$g(x) - \Delta y = f(x - \Delta x)$$

Seeing It seems simpler to just follow the evaluation of the function. For $g(x) = f(x + a) + b$, first you add $a$, then evaluate $f$ there, and finally add $b$. There's an extra layer of cognition when interpreting this as a translation of $f$ to $g$, because it entails a change in frame of reference (from $g$'s POV to $f$'s POV).

I think my confusion of this comes from how it was taught: instead of begining with function evaluation and then how it can be seen as a translation, we were taught the tramslation as a thing in itself, using the "rule" above. An "explanation" was given as an afterthought, secondary to the "rule". I'm not convinced the teacher had any real understanding beyond that, so they couldn't pass on an underetanding beyond the rule.

• The graph of $g(x)=f(x-a)$ is the set of points of the form $(x,f(x-a))$, while the graph of $f(x)$ is the set of points of the form $(x,f(x))$. But the former set is precisely the same as the set of points of the form $(x+a,f(x))$, which are just the points on the graph of $f(x)$ shifted $a$ units to the right. (By contrast, the points on the graph of $g(x)=f(x)+a$ are of the form $(x,f(x)+a)$, which are just the points $(x,f(x))$ shifted $a$ upward.) Jun 9, 2018 at 5:48
• @symplectomorphic great approach, using a more formal description of "shift". Can I request you write it as an answer, so I can accept it? I found it intimidating at first. I think it would ease a reader into the concept, by starting first with the graph of f(x) is the set of points of the form (x,f(x)), and omitting the vertical translation part, just to keep it simple (even though it's true). I've long been confused by this, so an extremely simple and direct answer would really help. Sorry to be so picky and annoying! Jun 10, 2018 at 5:39
• Here's my "Function Monkey" description from a few years back. (See here for an introduction to the Function Monkey himself.) After all this time, I see that I should probably streamline it (and perhaps add some actual images of coconuts) ... someday ...
– Blue
Jun 11, 2018 at 11:06
• @ symplectomorphic @ hyperpallium
– user655689
Mar 26, 2019 at 22:23
• @ symplectomorphic @ hyperpallium Thanks for all this! I think the key point is to understand why the set of points ( x , f(x-a) is identical to the set of points (x+a, f(x) ). Is there any simple algebraic proof of this? - I recently tried to answer the same question as this post and to provide a visual explanation, unfortunately I'm really bad at graphing as you'll see if you go to : " Is this visual explanation of horizontal shifting helpfull ( if not beautifull...)? "
– user655689
Mar 26, 2019 at 22:29

To keep it simple, lets assume $A>1,B>1,\Delta x>0$, and $\Delta y>0$. I like to think of the transformation $$g(x)=A\cdot f(B\cdot (x+\Delta x))+\Delta y\qquad (*)$$ as the composition $g(x)=V(f(H(x))$, where

• $H(x)=B\cdot(x+\Delta x)$ is a linear transformation done on the inputs before they are fed into $f$ (a so-called "horizontal" transformation), and
• $V(y)=A\cdot y+\Delta y$ is a linear transformation done on the outputs after they are produced by $f$ (a so-called "vertical" transformation).

More explicitly, given a set $\mathcal{D}\subset\mathbb{R}$, the set $H(\mathcal{D})=B\cdot(\mathcal{D}+\Delta x)$ is obtained by first shifting $\mathcal{D}$ to the right by $\Delta x$ to get the set $\mathcal{D}+\Delta x$, then enlarging that by a factor of $B$. Then the set $H(\mathcal{D})$ is passed into $f$, by which $f(H(\mathcal{D}))$ is produced. Finally, $$g(\mathcal{D})=V(f(H(\mathcal{D})))=A\cdot f(H(\mathcal{D}))+\Delta y$$ is obtained by enlarging $f(H(\mathcal{D}))$ by a factor of $A$ to obtain $A\cdot f(H(\mathcal{D}))$, then shifting that to the right by $\Delta y$ (notice I did not say "up" because we regard $f(H(\mathcal{D}))$ as a subset of $\mathbb{R}$).

So far everything has followed the order of operations, and $A,B>1$ correspond to enlargement and $\Delta x,\Delta y>0$ correspond to shifting in the positive direction. Note that in the special case that $f$ is invertible, one can write $g^{-1}$ in the same form as $(*)$: $$g^{-1}(y)=\frac{1}{B}\cdot f^{-1}\left(\frac{1}{A}\cdot(y-\Delta y)\right)-\Delta x.$$ As it should, this function applies all of the inverse steps to the set of $y$ values in the opposite order: shift the set of all $y$ values left by $\Delta y$, then shrink the set by a factor of $1/A$, then feed it through $f^{-1}$, shrink the image by $1/B$, then shift it left by $\Delta x$. This reveals the "symmetry" between the before/after linear transformations (this at least cleared up some confusion I used to have about why $A\cdot y+\Delta y$ and $B\cdot(x+\Delta x)$ were distributed differently).

However, I think the real source of the confusion is how all of this transformation is manifested on an $xy$-graph. I think this is because the graph compares the domain of $x$ values to the codomain of $y$ values, though the "symmetry" we saw above is actually between the domain and image of $f$. For example, we are familiar with the function $f(x)=\sin{x}$, its domain and image, and its graph. The way I was originally told to understand the graph of, say, $$g(x)=4\sin{3(x+2)}+1,$$ was to apply this sequence of transformations to the graph of $f$:

1. Horizontally shrink the graph by a factor of $\frac{1}{B}=\frac{1}{3}$,
2. Shift the graph left by $\Delta x=2$,
3. Vertically enlarge the graph by a factor of $A=4$,
4. Shift the graph up by $\Delta y=1$.

Though this gives the correct pictorial result, the "true" sequence is

1. Shift the domain right by $\Delta x=2$,
2. Then stretch the domain by a factor of $B=3$, (now feed this transformed domain into $f$)
3. Vertically enlarge the image of $f$ by a factor of $A=4$,
4. Shift the image up by $\Delta y=1$.

In both sequences, the last two steps (the vertical transformations) are the same because both represent transformations of the image of $f$ (not the codomain!). Steps 1. and 2. in the first sequence obscure/hide the fact that it is the domain that is being changed. In other words, the images of the two functions $f(x)=\sin(x)$ and $f(H(x))=\sin(3(x+2))$ are exactly the same, though their domains have been transformed.

So, very long story short: Think in terms of domain and image.

Exercise: Find and compare the fibers $f^{-1}(1)$ and $(f\circ H)^{-1}(1)$. How are these fibers explicitly related by $H$?

First note that if you think of $g(x)$ as $y$, then you've written

$$y-\Delta y = f(x-\Delta x).$$

So when you put the increments next to their parent variables, both are, in fact, subtracted.

Second, as to why subtracting moves things to the right, well, you've slowed $x$ down by $\Delta x$, so everything $f(x)$ does, $f(x-\Delta x)$ does later and "later" means further to the right.