Question

select all of the ordered pairs that make the inequality $y > x^2 + 3x - 4$ true.$square$ (0, 0)$square$ (2, 1)$square$ (-2, -1)$square$ (-5, -1)done

Explanation:

Step1: Test (0,0) in inequality

Substitute $x=0, y=0$:
$0 > 0^2 + 3(0) - 4$
$0 > -4$ (True)

Step2: Test (2,1) in inequality

Substitute $x=2, y=1$:
$1 > 2^2 + 3(2) - 4$
$1 > 4 + 6 - 4$
$1 > 6$ (False)

Step3: Test (-2,-1) in inequality

Substitute $x=-2, y=-1$:
$-1 > (-2)^2 + 3(-2) - 4$
$-1 > 4 - 6 - 4$
$-1 > -6$ (True)

Step4: Test (-5,-1) in inequality

Substitute $x=-5, y=-1$:
$-1 > (-5)^2 + 3(-5) - 4$
$-1 > 25 - 15 - 4$
$-1 > 6$ (False)

Answer:

(0, 0), (-2, -1)