I think the following is more in line with what your teacher has in mind:
_____________
y = √2x² + 5x - 12
2x² + 5x - 12 ≧ 0
(x + 4)(2x - 3) ≧ 0
Now we find critical numbers, which are the
zeros of the left side, by setting each factor
equal to 0:
x + 4 = 0; 2x - 3 = 0
x = -4; x = 3/2 = 1 1/2
We place these critical number on a number line:
<========⚫--------------------⚫--------------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
We use darkened circles instead of open circles because the
critical numbers are themselves solutions in this case since
the inequality is ≧ rather that >.
There are three intervals to test for solutions to the
inequality, the intervals between and beyond the critical
numbers
(-oo,-4), (-4, 3/2), and (3/2,oo). Or if you haven't had
interval notation, then they are these inequalities:
x ≦ -4, -4 ≦ x ≦ 3/2, x ≧ 3/2
Choose a test point in the interval (-oo,-4), say -5.
Substitute 5 into the inequality:
(x + 4)(2x - 3) ≧ 0
(-5 + 4)(2(-5) - 3) ≧ 0
(-1)(-10 - 3) ≧ 0
(-1)(-13) ≧ 0
13 ≧ 0
That is true, so (-oo,-4) is part of the solution, so we
shade that interval on the number line:
Choose a test point in the interval (-4,3/2), say 0.
Substitute 0 into the inequality:
(x + 4)(2x - 3) ≧ 0
(0 + 4)(2(0) - 3) ≧ 0
(4)(0 - 3) ≧ 0
(4)(-3) ≧ 0
-12 ≧ 0
That is false, so (-4,3/2) is NOT part of the solution, so we
DO NOT shade that interval on the number line, so we still have
jus:t
<========⚫--------------------⚫--------------
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
Choose a test point in the interval (3/2,-oo), say 2.
Substitute 2 into the inequality:
(x + 4)(2x - 3) ≧ 0
(2 + 4)(2(2) - 3) ≧ 0
(6)(4 - 3) ≧ 0
(6)(1) ≧ 0
6 ≧ 0
That is true, so (3/2,oo) is part of the solution, so we
shade that interval on the number line:
<========⚫--------------------⚫==============>
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
So the domain written in interval notation is
(-oo,-4] U [3/2, oo)
or as an inequality the domain is written this way:
x ≦ -4 OR x ≧ 3/2
Edwin