Function graph transformations 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?
 A: 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 $a$.
A: 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 $D+a$.
A: (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 $f(-(x-a))=f(a-x)$.
