I need help with this translation problem (pre-calc) I am having trouble understanding what I have to do with translation problems. I don't see anything in my textbook about it, and I can't find any resources online about it. Any help would be appreciated. Thanks. 
Here is the problem: 
The graph of $g(x)$ contains the point $(−5,6)$. Write a formula for a translation of $g$ whose graph contains the point: 
$(a) \ \  (-4, 6)$ then $y = $ ?
$(b) \ \ (-5, 8)$ then $y = $ ?
 A: Just to put what nsanger said into an answer, a translation of a function $g(x)$ will have the form $g(x) + c$ which is a vertical shift or the form $g(x + c)$ which is a horizontal shift.
So we see that the graph of $g(x)$ has the point $(-5, 6)$ and for $(a)$ we want the graph of $g(x)$ to have $(-4, 6)$.  So we are moving the $x$-coordinate $1$ unit to the right and so we are looking at a horizontal shift.  One tricky thing about horizontal shifts is that they are sometimes "reversed of what you might expect.  We can see that the difference between $-4$ and $-5$ is $+1$ so we might think that the shift is $g(x+1)$ but this actually shifts everything left, we want a shift right, which would be $g(x-1)$.  (One way to remember this, is to think that you are solving inside the parenthesis for $0$ so the first one gives $x+1=0 \Rightarrow x = -1$ so we are shifting left by $1$, while the second gives $x-1=0 \Rightarrow x =+1$ which is a shift right by $1$ which is what we wanted. In general, $g(x + c)$ moves the graph left $c$ units while $g(x - c)$ moves the graph right $c$ units (for positve $c$).
Now, for $(b)$ we see that the $x$-coordinate has not changed, but the $y$-coordinate has moved up by $2$.  So this is a vertical shift up $2$ units.  Vertical shifts act as we would expect, i.e. $g(x) + c$ moves the graph up $c$ units, but $g(x) - c$ moves the graph down $c$ units (for positive $c$).  So we can see that the translation we want is $g(x) + 2$.
