Why are horizontal transformations of functions reversed?

Question

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 ${}\times \frac{1}{a}$
• $f(x) + a$ - not grouped with $x$, vertical translation, shifts $y$-coordinate up, $a$
• $af(x)$ - not grouped with $x$, vertical scaling, $y$-coordinate ${} \times a$

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

Thanks for your help.

Answer 1

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.

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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.

Answer 2

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): $$\begin{align} x&\mapsto\frac{x-h}{a} \\ x_{\text{old}}&=\frac{x_{\text{new}}-h}{a} \\ ax_{\text{old}}&=x_{\text{new}}-h \\ ax_{\text{old}}+h&=x_{\text{new}} \end{align}$$

Answer 3

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