Find the point on the curve y = x³ - 11x + 5 at which the tangent is y = x - 11.

Find the point on the curve y = x³ - 11x + 5 at which the tangent is y = x - 11.

• A. (1, -5)
• B. (2, -9)
• C. (-2, 19)
• D. (3, -1)

Answer: B) (2, -9)

Explanation: Slope of given tangent line is 1. For curve, dy/dx = 3x² - 11. Equating slopes: 3x² - 11 = 1 → x² = 4 → x = ±2. At x = 2, y = 8 - 22 + 5 = -9. Point (2, -9) satisfies y = x - 11.