Stuck on what to put on the x and y table for this equation

Step-by-step explanation:

First off, let's take a look at the basic function here:  the absolute value function:  f(x) = |x|.  The graph of this function is v-shaped with vertex at (0,0).  On the right side of the y-axis, the function looks and acts like y = x (slope is +1).  On the left side of the y-axis, the function looks and acts like y = -x (slope is -1).   Please become familiar with this function, and make the effort to memorize these facts.

Now, given y = |x| and the determination to translate the vertex of the graph c units to the right, this y = |x| becomes y = |x-c|; the vertex is now at (c, 0).

If move the vertex c units to the left of (0,0), then y = |x| becomes y = |x+c|, and the vertex is at (-c, 0).  

If you move the vertex upward from (0,0), the function becomes y = |x| + d; if downward, y = |x| - d.

You can fill out that table if you want, but it's faster to use the info we just mentioned.  You are given y = |x - 3| - 1, whose vertex is at (3, -1).  So you immediately have one point to enter into the table.  You might next find the y-intercept; to do this, let x = 0 in y = |x - 3| - 1; you'll get |-3| - 1, which equals 2.  Your new point, the y-intercept, is (0, -2)

Lastly, let's find the x-intercepts.  At these, y = 0.  So, we set y = |x - 3| - 1 = 0.  Then |x - 3|= 1.  This tells us that x could be either 4 (remember that 4-3=1) or 2 (remember that 2-3 is -1 and |-1| = 1).  So we now have 2 more points:  (4,0), (2,0).

Your first entries in to that table could be x= 2, y = 0, and then x = 4, y = 0.  Several other points have been found in this discussion; write them into the table. Good luck!