Find the values of a and b so that the line 2x3ya is tangent

Solution

The curve passes through (1, -1) So, f(1) = -1 a + b + c = -1 --------------------- (1) f\'(x) = 3ax² + 2bx + c Slope of tangent there is -3, so. f\'(1) = -3 3a + 2b + c = -3 -------------- (2) f\'\'(x) = 6ax + 2b x = 1 is a point of inflection, so f\'\'(1) = 0 6a + 2b = 0 ---------------- (3) From (3), you get b = -3a. Substitute this value of b in (1) and (2) to get a and c. (1) -------------> c - 2a = -1 (2) -------------> c - 3a = -3 (2) - (1) -> a = 2 ---> b = -6 From (1), a = 2 ---> c = 3 So, a = 2, b = -6, c = 3