Question

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

Explanation:

Step1: Test (0, 0)

Substitute \( x = 0 \), \( y = 0 \) into \( y > x^2 + 3x - 4 \):
\( 0 > 0^2 + 3(0) - 4 \)
\( 0 > -4 \), which is true.

Step2: Test (2, 1)

Substitute \( x = 2 \), \( y = 1 \) into \( y > x^2 + 3x - 4 \):
\( 1 > 2^2 + 3(2) - 4 \)
\( 1 > 4 + 6 - 4 \)
\( 1 > 6 \), which is false.

Step3: Test (-2, -1)

Substitute \( x = -2 \), \( y = -1 \) into \( y > x^2 + 3x - 4 \):
\( -1 > (-2)^2 + 3(-2) - 4 \)
\( -1 > 4 - 6 - 4 \)
\( -1 > -6 \), which is true.

Step4: Test (-5, -1)

Substitute \( x = -5 \), \( y = -1 \) into \( y > x^2 + 3x - 4 \):
\( -1 > (-5)^2 + 3(-5) - 4 \)
\( -1 > 25 - 15 - 4 \)
\( -1 > 6 \), which is false.

Answer:

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