Interpreting graph transformation problems I seem to get really confused with these simple graph translation problems in words. For example, the function $f(x)=2x^2+3$ is translated 3 units in the positive direction parallel to the x-axis. I interpreted movement parallel to the x-axis as a vertical translation, and so I thought the transformed f(x) would just be $f(x)=2x^2+6$, but apparently it's $(2x-3)^2+3$.
Also, the graph of the function  g(x) obtained from the graph of $f(x)=xe^x$ by stretching it in the x-direction by a scale factor 2. What exactly does stretching in the "x-direction" mean? 
 A: To translate a function $f(x)$, simply do $f(x-a)$, where $a$ is a translation of $a$ units parallel to the $x$-axis. From this we get $$f(x-3) = 2(x-3)^2+3$$ Which is equivalent to $2(x-3)^2+3$
To stretch a function in the $x$-direction by a factor of $c$, we multiply $x$ by $\frac{1}{c}$. From this we see $$g(x) = \frac{1}{2}xe^{\frac{1}{2}x} \longrightarrow g(x) = \frac{xe^{\frac{x}{2}}}{2}$$
A: I think that it helps to remember you are really transforming the axis not the graph.  
An anolgy that I use is the table cloth trick whereby someone yanks a tablecloth and all the cups and saucers remain exactly where they were.  
In your example the in order to get the graph to appear to move to the right you need to pull the $x$ axis by $3$ units to the left.   The $x$ axis is the tablecloth and the graph the cups and saucers. 
More specifically $x \mapsto x-3$.
So $2x^2+3 \to 2(x-3)^2+3$ as I've mentioned in the comments.
Stretching is slightly different.
To stretch by a scale factor 2.
You need to use $x \mapsto \frac{1} {2} x$. 
Why? Imagine the last $x$ value to be plotted before the transformation, this completes the graph. After the transformation this will only be the half way point of the graph.
More specifically $xe^{x} \to \frac{1} {2}x e^{1/2x}$
A: Moving it parallel to the $x$ axis along he positive $x$ means that you are moving the graph to the right by some given amount. Generally you can think of it as moving the point at which the function has it $x =0 $.
If a function is moved the along the $x$ axis; it' ll be done like this;
$f(x) \to f(x\pm a)$ where $a$ is the amount you translate the given function. If moving to the right it becomes $f(x-a)$ if moving to the left it is $f(x+a)$.
Similarly  moving it up and down means you are changing the point where the function has $y = 0$.
Here $f(x) \to f(x)\pm a$ where $a$ is the amount by which you translate along the $y$ axis.If moving up it becomes $f(x)+2$ and moving down results in $f(x)-a$
For example take $\sin(x)$ and let us translate it by $2$ units in both the $x$ and $y$ axes separately. 
Here is the original $y=\sin(x)$
Translating it to the left by $2$ gives;

Translating it up by $2$ units gives;

Scaling is similar to it, you can visualize it by thinking of either pulling the graph apart by some factor or squishing it by some factor;
When squishing by some factor $a$ along the $x$ axis
; $f(x) \to f(ax)$ when streching by $a$ along $x$ axis ; $f(x)\to f(\frac xa)$ 
when Stretching  or squeezing by $a$ along the $y$ axis ; $f(x) \to af(x)$ for streaching and $f(x) \to\frac 1af(x)$  for squeezing.
For example take $y= x\cdot e^x$
originally ;

squeezing along x;

and stretching along x;

Experiment transforming functions and you will begin to under how these transformations work.
