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.