question_answer 1)

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

Answer:

Given curve is             y = x3 ? 11x + 5             ?.(1)       Let P(h, k) be a point on (1)          K = h2 ? 11 x + 5           ?. (2)       (1)       Given that slope of tangent to (1) = 1                               When h = 2,       (2)  K = 8 ? 22 + 5 = ?9         When h = ?2,                 (2)  K = ?8 + 22 + 5 = 19       Point on (1) are (2, ?9) and (?2, 19)       Now equation of tangent to curve (1)       At (2, ?9) y ? k = m (x ? h)       y + 9 = 1 (x ? 2)       y = x ? 11       At (?2, 19)         y ? 19 = 1 (x + 2)        y = x + 21.       Hence the point on (1) where the tangent to (1) is       y = ?11, is (2, ?9).