y > 2x² - 6x - 36 is the inequality in standard form represents the region greater than the quadratic function with zeros –3 and 6 and includes the point (–2, –16) on the boundary line. This can be obtained by finding zeroes of each inequality to check whether the zeroes are –3 and 6 and then substituting (–2, –16) in the inequality.
<h3>Find the required inequality:</h3>
To find the zeroes of the inequality we use the formula,
x = (-b ± √b² - 4ac)/ 2a, where a, b and c are the coefficients of x², x and constant respectively.
- Option 1 : y > 1/2 x² - 3/2x - 9
Find the zeroes,
x = (3/2 ± √9/4 + 18)
x = 3/2 ± 9/2
⇒ x = 12/2 = 6, x = -6/2 = - 3
Putting x = –2 in the inequality we get,
1/2 x² - 3/2x - 9 = 1/2 (-2)² - 3/2(-2) - 9
= 2 + 3 - 9 = - 4 ≠ -16
Therefore option 1 is incorrect
- Option 2 : y > 2 x² + 6x - 36
Find the zeroes,
x = (-6 ± √36 + 288) / 4
x = (-6 ± 18) / 4
⇒ x = 12/4 = 3, x = -24/4 = - 6
Therefore option 2 is incorrect
- Option 3 : y > 2 x² - 6x - 36
Find the zeroes,
x = (6 ± √36 + 288) / 4
x = (6 ± 18) / 4
⇒ x = 24/4 = 6, x = -12/4 = - 3
Also , Putting x = –2 in the inequality we get,
2 x² - 6x - 36 = 2 (-2)² - 6(-2) - 36
= 8 + 12 -36
= - 16
The inequality includes the point (–2, –16) on the boundary line.
Hence y > 2x² - 6x - 36 is the inequality in standard form represents the region greater than the quadratic function with zeros –3 and 6 and includes the point (–2, –16) on the boundary line.
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Question: Which inequality in standard form represents the region greater than the quadratic function with zeros –3 and 6 and includes the point (–2, –16) on the boundary line?
a) y > 1/2 x² - 3/2x - 9
b) y > 2 x² - 6x - 36
c) y > 2 x² - 6x - 36
d) y > 1/2 x² + 3/2x - 9
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