Question 1130516
If this is 4/(x+2) then there is a vertical asymptote at x=-2, because that makes the denominator 0.
the y-intercept is at (0, 2)
there is a horizontal asymptote at y=0, approaching from the positive side at positive x and from the negative side with negative x. 
The range is from -oo to +oo,  The domain is all x except x= -2

{{{graph(300,300,-10,10,-10,10,4/(x+2))}}}

If it is (4/x)+2, then it has a vertical asymptote at x=0, it is displaced upwards 2 units, making the horizontal asymptote +2 and the same general shape, shifted upward 2 and to the right 2.
{{{graph(300,300,-10,10,-10,10,(4/x)+2)}}}