What represents the inequality |x + 1| + 2 < –1.

Question is a bit vague.  If you wish to graph this inequality, you'll need to know what the graph of the absolute value function y = |x| looks like; it's a " V " with the vertex at the origin.  The slope of the right half of the graph is m=1.  Draw this function.  

Next, subtract 2 from both sides.  We'll get  |x + 1|  < –1 - 2
or 
 |x + 1|  < –3.  We can stop here!  Why!  because the absolute value function is never smaller than zero, and so  |x + 1| is never smaller than -3.  

You could, of course, graph y = |x+1|; start with your graph for y = |x| and then move the whole graph 1 unit to the left (away from the origin).  If you do this properly you'll see that the entire graph is above the x-axis, except for the vertex (-1,0).  Again, that tells us that the given inequality has no solution.

Aline graph should be used