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I'm looking for a reliable document with formal mathematical justification of how and why graph transformations work. Given a function $f(x)$, what is the graph of $f(a-x)$? Why should we shift $f(x)$ by $a$ to the right in this case? Imagine someone read a paper about graph transformations, like this one.

Intuitively, someone could say: well, $f(a-x)$ is a graph of $f(-x)$ shifted left by $a$ because we are adding $a$, right? Is there a more elegant and formal way to prove what those transformations to graphs result in?

user4205580
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3 Answers3

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Define the graph of a function $f:D\rightarrow\mathbb R$ to be the set of points $$ G(f)=\{(x,f(x))\ \mid\ x\in D\} $$ Then the function $g(x)=f(x-a)$ is defined whenever $x-a=y\in D$ so that $x\in E=\{y+a\ \mid y\in D\}$ and the graph is $$ \begin{align} G(g)&=\{(x,g(x))\ \mid\ x\in E\}\\ &=\{(x,f(x-a))\ \mid\ x\in E\}\\ &=\{(y+a,f((y+a)-a))\ \mid y\in D\}\\ &=\{(y+a,f(y))\ \mid y\in D\} \end{align} $$ So when you compare this last description of $G(g)$ to the description of $G(f)$ you see that $$ (x,f(x))\in G(f)\iff(x+a,f(x))\in G(g) $$ So the graph of $g$ contains all the same points as the graph of $f$ with the $x$-value of each point shifted by $a$.

UPDATE: In particular this means that for the function $h(x)=f(-x)$ we have $h(x-a)=f(-(x-a))=f(a-x)$ which has the same graph as $h$, but with the $x$-value of every point shifted by $a$.

String
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  • Is it ok that you changed $y$ to $x$ here?: ${(y+a,f(y))\ \mid y\in D}$ and then you got: $(x,f(x))\in G(f)\iff(x+a,f(x))\in G(g)$. – user4205580 Jan 17 '15 at 22:03
  • @user4205580: Indeed! Both $x$ and $y$ belongs to the domain $D$ so they are really both names of variables to range over $D$ and produce the graphs $(x,f(x))$ and $(x+a,f(x))$ respectively. – String Jan 17 '15 at 22:36
  • Oh, I just remembered something. Is this statement correct? "The $x$-coordinates in the graph of $y=f(x)$ correspond to the $x$-coordinates in the graph of $y=f(cx)$ multiplied by $c$." I believe it should be the other way round. – user4205580 Feb 06 '15 at 17:06
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The graph of $x\mapsto f(x+a)$ is just the composition of $f$ with the translation $x\mapsto x+a$. Therefore, if $D$ is the domain of $f$, the graph of $x\mapsto f(x+a)$ is the same graph than $f$ but on the domain $D+a$.

idm
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  • Ok. How would you convince me that the graph of $f(a-x)$ is obtained by shifting $f(-x)$ to the right by $a$? – user4205580 Jan 17 '15 at 19:56
  • Why don't you draw the two functions ? take for exemple $f(x)=x$ or $f(x)=x^2$, you'll see immediately ! But actually, I don't have to convince you ! it's just logical ! – idm Jan 18 '15 at 00:07
  • You really think I can't see it? My point is how would you convince a person who can't. – user4205580 Jan 18 '15 at 16:55
  • to make a draw, it's the easier way to explain it :-) – idm Jan 18 '15 at 17:39
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(Thinking of $a>0$, as it seems that you do)

You seem to be OK (correct me if I'm wrong) that $f(x-a)$ shifts the graph of $x\mapsto f(x)$ to the right with $a$ steps. OK?

Thus, to shift the function $x\mapsto f(-x)$ to the right $a$ steps, we want $f(-(x-a))=f(a-x)$.

mickep
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  • Ok, but how do conclude that if $f(x-a)$ is shifted $a$ to the right, then $f(-(x-a))$ will shift the graph to the right as well? The purpose of this thread is to prove it with as little referring to intuition as possible. – user4205580 Jan 17 '15 at 21:31
  • @user4205580: I just updated my answer. In my update, I believe I gave the justification formally for what you originally asked. I hope you agree! – String Jan 17 '15 at 21:51