$y = f(x)$ $y = f(x) + 2\,$ up $2$ 
$y = f(x)$ $y = f(x)  2\,$ down $2$ 
$y = f(x)$ $y = f(x+2)\,$ left $2$ 
$y = f(x)$ $y = f(x2)\,$ right $2$ 
movements up and down change the $y$values of points; transformations that affect the $y$values are intuitive (e.g., to move up $\,2\,$, you add $\,2\,$ to the previous $\,y\,$value) 
movements left and right change the $x$values of points; transformations that affect the $x$values are counterintuitive (e.g., to move left $\,2\,$, you replace every $\,x\,$ by $\,x+2\,$, NOT $\,x2\,$) 

Shifting up/down/left/right does NOT change the shape of a graph. 
The lesson
Graphing Tools: Vertical and Horizontal Translations
in the
Algebra II curriculum
gives a thorough discussion of shifting graphs up/down/left/right.
The key concepts are repeated here.
The exercises in this lesson duplicate those in
Graphing Tools: Vertical and Horizontal Translations.
Start with the equation
$\,\color{purple}{y=f(x)}\,$.
Adding $\,\color{green}{p}\,$ to the previous $\,\color{green}{y}\,$values gives the new equation $\,\color{green}{y=f(x)+p}\,$. This shifts the graph UP $\,\color{green}{p}\,$ units. A point $\,\color{purple}{(a,b)}\,$ on the graph of $\,\color{purple}{y=f(x)}\,$ moves to a point $\,\color{green}{(a,b+p)}\,$ on the graph of $\,\color{green}{y=f(x)+p}\,$ . 

Additionally:
Start with the equation $\,\color{purple}{y=f(x)}\,$. Subtracting $\,\color{green}{p}\,$ from the previous $\,\color{green}{y}\,$values gives the new equation $\,\color{green}{y=f(x)p}\,$. This shifts the graph DOWN $\,\color{green}{p}\,$ units. A point $\,\color{purple}{(a,b)}\,$ on the graph of $\,\color{purple}{y=f(x)}\,$ moves to a point $\,\color{green}{(a,bp)}\,$ on the graph of $\,\color{green}{y=f(x)p}\,$. 



Notice that different words are used when talking about transformations involving $\,y\,$, and transformations involving $\,x\,$.
For transformations involving
$\,y\,$
(that is, transformations that change the $\,y$values of the points),
we say:
vertical translations:
going from
$\,y=f(x)\,$
to
$\,y = f(x) \pm c\,$
horizontal translations:
going from
$\,y = f(x)\,$
to
$\,y = f(x\pm c)\,$
Make sure you see the difference between (say) $\,y = f(x) + 3\,$ and $\,y = f(x+3)\,$!
In the case of
$\,y = f(x) + 3\,$, the $\,3\,$ is ‘on the outside’;
we're dropping $\,x\,$ in the $\,f\,$ box, getting the corresponding output, and then adding $\,3\,$ to it.
This is a vertical translation.
In the case of
$\,y = f(x + 3)\,$, the $\,3\,$ is ‘on the inside’;
we're adding $\,3\,$ to $\,x\,$ before dropping it into the $\,f\,$ box.
This is a horizontal translation.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. Note: There are lots of questions like this: “Start with $\,y = f(x)\,$. Move UP $\,2\,$. What is the new equation?” Here is the same question, stated more precisely: “Start with the graph of $\,y = f(x)\,$. Move this graph UP $\,2\,$. What is the equation of the new graph?” All the ‘graphs’ are implicit in the problem statements. (When something is implicit then it's understood to be there, even though you can't see it.) 
PROBLEM TYPES:
