|y| = x + 1
Since |y| ≧ 0, x + 1 ≧ 0, hence x ≧ -1, so we cannot use any values
of x less than 1.
So we get a few points.
When x = -1 (the smallest value x can take on)
|y| = -1 + 1
|y| = 0
y = 0, so we have the point (1,0) as the left-most point on the graph.
When x = 0,
|y| = 0 + 1
|y| = 1
There are two values of y that satisfy that, 1 and -1,
so we have the two points (0,1) and (0,-1)
When x = 0,
|y| = 0 + 1
|y| = 1
There are two values of y that satisfy that, 1 and -1,
so we have the two points (0,1) and (0,-1)
When x = 1,
|y| = 1 + 1
|y| = 2
There are two values of y that satisfy that, 2 and -2,
so we have the two points (1,2) and (1,-2)
That's enough.
We plot what we have:
Then draw the graph:
Edwin
