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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?

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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 $$f(x)\to f(x+3)\to f(-x+3)\to f(-(x-3)+3)=f(6-x)$$

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  • $\begingroup$ How would the transformation look like if I would have to reflect the graph $y=f(x)$ about a line parallel to the x-axis, e.g. $y=3$? You would have to shift down by 3, reflect, and shift up again. I would then have $$y\to y-3\to -y-3\to -(y+3)-3 = -y-6$$, as $y-3$ shifts down and $y+3$ shifts up. This is wrong, so where is my mistake this time? $\endgroup$ – FizzleDizzle Oct 7 '18 at 9:12
  • $\begingroup$ @FizzleDizzle $y$-axis transformations affect the entire expression (i.e. they are applied after applying the meat of the function to be transformed, in contrast to $x$-axis transformations, which are applied before). $\endgroup$ – Parcly Taxel Oct 7 '18 at 11:44
  • $\begingroup$ Ah ok. So $$y\to y-3\to -y+3\to -y+3+3 = -y+6$$ would be correct? $\endgroup$ – FizzleDizzle Oct 7 '18 at 12:37
  • $\begingroup$ @FizzleDizzle Yes. $\endgroup$ – Parcly Taxel Oct 7 '18 at 12:38
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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 $y=f(6-x).$

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