<p>What is the minimum height of any point on the curve \(y = x^2 - 4x + 6\) above the \(x\)-axis?</p>
Step-by-Step Solution
Key Concept: The minimum value of a quadratic y = ax² + bx + c occurs at the vertex x = -b/(2a), and since a > 0, the parabola opens upward ensuring this is a global minimum.
<p><strong>Step 1:</strong> Identify the quadratic form y = x² - 4x + 6, where a = 1, b = -4, c = 6.</p><p><strong>Step 2:</strong> Find the x-coordinate of the vertex using x = -b/(2a) = -(-4)/(2·1) = 4/2 = 2.</p><p><strong>Step 3:</strong> Substitute x = 2 into the equation to find the minimum y-value:<br/>y = (2)² - 4(2) + 6 = 4 - 8 + 6 = 2</p><p><strong>Step 4:</strong> Verify by completing the square: y = x² - 4x + 6 = (x - 2)² - 4 + 6 = (x - 2)² + 2. Since (x - 2)² ≥ 0, the minimum value is 2 when x = 2.</p><p>∴ <strong>Answer: 2</strong></p>
Correct Answer: 2