Shifting the graph of a function

Question

I have a conceptual problem.

If the graph of a function $f(x)$ has to be horizontally shifted to the right by $a$ units, then the new graph is $f(x-a)$.

However, if the graph has to be reflected about a vertical line, e.g. $x=3$, then the graph needs to be translated by $3$ units to the left, reflected and shifted back to the right by $3$ units (as described here). According to the rule above, the new graph should be $$f(x) \rightarrow f(x+3) \rightarrow f(-x-3) \rightarrow f(-x-3-3)=f(-x-6)$$ This is wrong, and the right answer $f(6-x$) can only be achieved by switching the operations $±3$ for translating to the left and to the right.

Could someone enlighten me?

Answer 1

Shifting right by $a$ units is a substitution $x\to x-a$. This substitution does not affect anything outside the $x$'s. Similarly, reflection about the $y$-axis is a substitution $x\to-x$, again not affecting other symbols.

Thus the reflection about $x=3$ proceeds $$f(x)\to f(x+3)\to f(-x+3)\to f(-(x-3)+3)=f(6-x)$$

Answer 2

Let $x'$ be a reflection of $x$ wrt. the axis $x=3$. Then the arithmetic mean of $x$ and $x'$ is $3$, so $\frac{x+x'}{2}=3$, hence $x'=6-x.$ Therefore the image of the curve $y=f(x)$ under this reflection is $y=f(6-x).$