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While studying graph transformations I came across horizontal and vertical scale and translations of functions. I understand the ideas below.

  • $f(x+a)$ - grouped with x, horizontal translation, inverse, x-coordinate shifts left, right for -a
  • $f(ax)$ - grouped with x, horizontal scaling, inverse so x-coordinate * 1/a
  • $f(x)$ + a - not grouped with x, vertical translation, shifts y-coordinate up, d
  • $af(x)$ - not grouped with x, vertical scaling, y-coordinate * a

I have mostly memorized this part but I am unable to figure out why the horizontal transformations are reversed/inverse?

Thanks for your help.

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For $x=-a$ $f(x+a)=f(0)$ and for $x=a$ $f(x-a)=f(0)$. –  Américo Tavares Jun 18 '11 at 9:58
    
Sorry, I am not sure what you are saying. But thanks anyway. @Jonas's explanation helped clear things up nicely. –  mathguy80 Jun 18 '11 at 16:59
    
It might help to remember the following mantra: "f(x + a) makes 0 a, and f(x - a) makes a 0." –  Hexagon Tiling Jan 7 '12 at 11:14

3 Answers 3

up vote 5 down vote accepted

You're really talking about what happens to the graph $y=f(x)$; and from this perspective, we can see that horizontal (x) and vertical (y) transformations work the same way.

Instead of writing $y=f(x)+a$, write $y+b=f(x)$ (here, $b=-a$); and instead of $y=af(x)$, write $by=f(x)$ (here, $b=\frac{1}{a}$).

So for translations, we have

  • $y=f(x+a)$ shifts $x$ by $-a$.
  • $y+b=f(x)$ shifts $y$ by $-b$.

And for scaling, we have

  • $y=f(ax)$ scales $x$ by $1/a$.
  • $by=f(x)$ scales $y$ by $1/b$.

So you see, they really work the same way, it just looks opposite because the factor $a$ gets moved to the other side.


This works very generally. Suppose we have

  • equation 1: $F(x,y,z)=0$,
  • equation 2: $F(x,y,z+c)=0$, and
  • equation 3: $F(x,y,dz)=0$.

Now take any solution to equation 1, lets call it the triple $(n_1,n_2,n_3)$. (So equation 1 is true if I plug in the numbers $n_1$ for $x$, $n_2$ for $y$, and $n_3$ for $z$.)

Then you can see that $(n_1,n_2,n_3-c)$ is a solution of equation 2, and $(n_1,n_2,\frac{1}{d}n_3)$ is a solution of equation 3.

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This is an excellent answer. You've clarified things and then some! Thank you! –  mathguy80 Jun 18 '11 at 16:52

Jonas Kibelbek's answer covers almost all of what I'd have said. The one thing I'd add is that the substitution $$x\mapsto\frac{x-h}{a}$$ (or similarly $y\mapsto\frac{y-k}{b}$) is a dilation by a factor of $a$ centered at $0$ (if $|a|>1$, it's a stretch; if $|a|<1$, it's a shrink), followed by a translation by $h$ (if $h>0$, in the positive direction (and similarly for $y$, $b$, and $k$). One way to think of this is to change the way we're writing the mapping a bit (still talking about the same mapping, just writing it differently): $$\begin{align} x&\mapsto\frac{x-h}{a} \\ x_{\text{old}}&=\frac{x_{\text{new}}-h}{a} \\ ax_{\text{old}}&=x_{\text{new}}-h \\ ax_{\text{old}}+h&=x_{\text{new}} \end{align}$$

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Nicely put! That helps too. Thank you, @Issac! –  mathguy80 Jun 18 '11 at 16:57

For the same reason scrolling down makes a document move up.

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To the downvoter: This was not meant to be a flip answer. I've used this analogy in teaching and it has proven to be enlightening one, in the right hands. I would ask that you give it a little more thought. –  I. J. Kennedy Oct 9 '11 at 18:15
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Kennedy: Your answer is a good one, and I have up-voted it. I would add that what you cite is an instance of the more general phenomenon that one gear turning another causes that second gear to go in the opposite direction, and, even more generally, an instance of the principle of action-and-reaction. –  Hexagon Tiling Jan 7 '12 at 19:21
    
@I.J.Kennedy I didn't downvote you (I'm just reading your answer now), but I think your answer would have less likely been interpreted as flip if you provided just a bit more detail. For example, what is the reason scrolling down makes a document move up? –  yroc Mar 7 at 17:17
    
@yroc Thanks for the comment. When I wrote my answer, here's what I was imagining: "...same reason scrolling makes a document go up" Huh? (Moves mouse a few inches to the right, tries scrollbar.) Oh! Of course. –  I. J. Kennedy Mar 7 at 19:54

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