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.
Everything looks correct except for the order in which the horizontal transformations are applied. The transformation
$$ y = x \qquad \leadsto \qquad y = \frac{1}{cx + d} $$
first shifts the graph left by $d$, then scales towards the $y$-axis by a factor of $c$, which means that the shift gets scaled as well.
If you like, factor: $cx + d = c\left(x + \tfrac{d}{c}\right)$, which allows you to apply the transformations in the usual order: shrink by factor of $c$, then shift by $-\tfrac{d}{c}$.
Best Answer
$$y=Af(B(x+\frac{C}{B}))+D$$
Can be thought of taking $f(x)=y$ and performing the following substitution.
$(x,y) \mapsto (Bx+C, \frac{y-D}{A})$
In order to understand what works and what doesn't work you need to understand what's going on.
Here is what is going on:
Let's say you have some function $y=f(x)$, it has some graph. This graph is a set $G$ consisting of points $(x,y)$ where $x$ is in the domain of the function.
If you consider $f(x,y)=y-f(x)=0$ then for every substitution you perform you'll witness an inverse mapping in the graph.
For example say we perform $x \mapsto x+1$, so now we have $y-f(x+1)=0$. You might expect the graph to be composed of points $(x+1,y)$ with respect to the old graph, but this is not true rather it is composed of points $(x-1,y)$, i.e. a shift left.
On the other hand say we perform $x \mapsto 2x$, now we have $y-f(2x)=0$. Now because the inverse of the mapping $x \mapsto 2x$ is $x \mapsto \frac{1}{2}x$ now the points become,
$$(\frac{1}{2}x,y)$$
Sometimes a combination of shifts, dilations, etc are needed, for example $y=x^2$ to $y=(2x+1)^2+1$ requires the substitution $(x,y) \mapsto (2x+1,y-1)$ whose inverse $(x,y) \mapsto (\frac{x-1}{2},y+1)$ tells you exactly what to do to the graph.
Computing the inverse of $(x,y) \mapsto (Bx+C, \frac{y-D}{A})$ will tell you everything you want to know.
I get $(x,y) \mapsto (\frac{x-C}{B},Ay+D)$. (You can perform this on points in your graph, one step at a time, in whichever way makes sense).
For example first shifting all $x$ coordinates to the left $C$, then scaling them by $\frac{1}{B}$, then scaling $y$ coordinates by $A$, then shifting up by $D$ makes sense.
But, doing all the same for $x$ and then shifting up $y$ by $D$ to get to $y+D$ then scaling by $A$ to get to $A(y+D)$ doesn't make sense!