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 * 1/a
- $f(x)$ + a - not grouped with x, vertical translation, shifts y-coordinate up, d
- $af(x)$ - not grouped with x, vertical scaling, y-coordinate * 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.
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$\begingroup$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.
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.
$\endgroup$ 1 $\begingroup$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): $$\begin{align} x&\mapsto\frac{x-h}{a} \\ x_{\text{old}}&=\frac{x_{\text{new}}-h}{a} \\ ax_{\text{old}}&=x_{\text{new}}-h \\ ax_{\text{old}}+h&=x_{\text{new}} \end{align}$$
$\endgroup$ 1 $\begingroup$For the same reason scrolling down makes a document move up.
$\endgroup$ 4 $\begingroup$For Horizontal transformations:
For example if you have y=(x-3), this would mean that you shifted the equation from y=x by 3 units right..
The way you want to think of this is that to get back to the point (0,0) from (3,0) is that you need to subtract 3. This is where the negative sign comes from and the reason why the sign is not + instead!
Reference:
$\endgroup$ $\begingroup$For horizantal shift: The logical reason for horizantal shift is that in(f)(x)=y=x the origin is (0,0)and in f(x)=(x-2)is (2,0) for this we should add 2 to get 0 becouse in parent function become 0 when we add 0 and in shifted function to make zero our function we ahould add 2
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