1. The domain of the relation {(x, y): y = 2x^2 + 1} is all real numbers.
To see why, note that the expression 2x^2 + 1 is valid for any real value of x. Therefore, there are no restrictions on the domain of the relation, and it includes all possible ordered pairs of the form (x, y), where y is defined by the expression 2x^2 + 1.
2. To complete the set of ordered pairs for the relation {(x, y): y = 2 |x + 1| and x ∈ {-2, -1, 0, 1, 2}}, we need to evaluate the expression 2|x + 1| for each value of x in the given set, and record the resulting value of y.
For x = -2, we have y = 2|-2 + 1| = 2*1 = 2, so the ordered pair is (-2, 2).
For x = -1, we have y = 2|-1 + 1| = 2*0 = 0, so the ordered pair is (-1, 0).
For x = 0, we have y = 2|0 + 1| = 2*1 = 2, so the ordered pair is (0, 2).
For x = 1, we have y = 2|1 + 1| = 2*2 = 4, so the ordered pair is (1, 4).
For x = 2, we have y = 2|2 + 1| = 2*3 = 6, so the ordered pair is (2, 6).
Therefore, the set of ordered pairs for the relation is {(-2, 2), (-1, 0), (0, 2), (1, 4), (2, 6)}.
So the answer to your question is (-2, 2).