Based on the diagram of $f(x)$ draw the $g(x)$ Based on the diagram of $f(x)$ draw the $g(x)$?
$f(x)$'s diagram : 
$g(x)= f(x/2 + 1)$

one solved it so: fist effected "$-1$" then "$*2$" to the "$x$"

the other one first effected "$*2$" then "$-1$" to the "$x$"

Now I don't know which one true, since both are two teachers, what I guess is that the "$*2$" must be applied sooner?
 A: The first answer is correct. For each pair of coordinates $(x,y)$ in the equation $y=f(x)$, there is a corresponding pair $(x',y)$ for the equation $y=f(x')$ where $x'=\frac{x}{2}+1$.
But we want to draw $y=f(x')$ on the $x-y$ plane, not the $x'-y$ plane, so we want to change the values of $x$ so that $x'$ takes the values $-1,0,0,2$ like in the black tables.
$x=2x'-2$ so we want to double each $x$ value, and then subtract 2, which is what the first teacher does.
An easy way to verify this is to plug in the values of the first red table into $f$ and see that they give the same values of $y$ as the black table does.
A: The graph of $g$ is the subset of the plane $\mathbb R^2$ given by
$$
G_g = \{(x,g(x)) \mid x \in \mathbb R\} = \{(x,f(x/2+1)) \mid x \in \mathbb R\} = \{((2x-2,f(x)) \mid x \in \mathbb R\}.
$$
We can write this more compactly as an affine transformation of the graph of $f$, namely $G_g  = A G_f + b$, where $A := \begin{bmatrix}2 & 0\\0 & 1\end{bmatrix}$, and $b := (-2,0)$. Thus,

The "diagram" of $g$ is the diagram of $f$ transformed as follows: you stretch the diagram in the $x$-direction by factor of $2$ and the translate the resulting diagram leftward $2$ units!

