Question

select the point that satisfies $y \leq x^2 - 4x + 2$.

a. (5, 5)

b. (2, 2)

c. (1, 1)

d. (4, 4)

Explanation:

Step1: Test Point A (5,5)

Substitute \(x = 5\) and \(y = 5\) into \(y\leq x^{2}-4x + 2\).
Calculate \(x^{2}-4x + 2\) when \(x = 5\): \(5^{2}-4\times5 + 2=25 - 20+2 = 7\).
Check if \(5\leq7\): Yes, but we need to check all points.

Step2: Test Point B (2,2)

Substitute \(x = 2\) and \(y = 2\) into \(y\leq x^{2}-4x + 2\).
Calculate \(x^{2}-4x + 2\) when \(x = 2\): \(2^{2}-4\times2 + 2=4 - 8 + 2=-2\).
Check if \(2\leq - 2\): No.

Step3: Test Point C (1,1)

Substitute \(x = 1\) and \(y = 1\) into \(y\leq x^{2}-4x + 2\).
Calculate \(x^{2}-4x + 2\) when \(x = 1\): \(1^{2}-4\times1 + 2=1 - 4 + 2=-1\).
Check if \(1\leq - 1\): No.

Step4: Test Point D (4,4)

Substitute \(x = 4\) and \(y = 4\) into \(y\leq x^{2}-4x + 2\).
Calculate \(x^{2}-4x + 2\) when \(x = 4\): \(4^{2}-4\times4 + 2=16 - 16+2 = 2\).
Check if \(4\leq2\): No. Wait, earlier for A, \(5\leq7\) is true. Wait, maybe I made a mistake. Wait, let's re - check point A: \(x = 5\), \(y = 5\). \(x^{2}-4x + 2=25 - 20 + 2 = 7\). \(5\leq7\) is true. But wait, let's check again. Wait, maybe the options: Wait, maybe I miscalculated for point A. Wait, the problem is to find the point that satisfies \(y\leq x^{2}-4x + 2\). Let's re - check each point:

For point A: \(x = 5\), \(y = 5\). \(x^{2}-4x + 2=25-20 + 2 = 7\). \(5\leq7\) is true.

For point B: \(x = 2\), \(y = 2\). \(x^{2}-4x + 2=4 - 8+2=-2\). \(2\leq - 2\) is false.

For point C: \(x = 1\), \(y = 1\). \(x^{2}-4x + 2=1 - 4 + 2=-1\). \(1\leq - 1\) is false.

For point D: \(x = 4\), \(y = 4\). \(x^{2}-4x + 2=16-16 + 2 = 2\). \(4\leq2\) is false.

So the only point that satisfies the inequality is (5,5).

Answer:

A. (5, 5)