Question

A curve passes through the point 1,-11 and its gradient at any point is ax2+b, where a and b are constants. The tangent to the curve at point 2,-16 is parallel to the x axis. Find the values of a and b.

Answer 1

Answer:

a = 3

b = -12

Step-by-step explanation:

gradient function (y') = ax² + b

curve function (y) = 1/3 ax³ + bx + c  

pass (1,-11) : -11 = 1/3 a + b + c  ..... (1)

pass (2,-16) : -16 = 8/3 a + 2b + c  ..... (2)

slope at (2,-16) is 0 (its tangent parallel to x axis)

at (2 , -16) ax² + b = 0         4a + b = 0         b = -4a   .... (3)

(2) - (1) : -5 = 7/3 a + b = 7/3 a -4a = -5/3 a   ... (4)

3 x (4) : -15 = -5a            a = 3

b = -4a = -4 x 3 = -12