Question from http://www.sgforums.com/forums/2297/topics/343229
A curve has the equation y = 3x3 - 5x + 4
(a) Find the equation of the tangent to the curve at A(1,2)
(b) Find the coordinates of the other point on the curve at which the tangent is parallel to the tangent at A.
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Answer:
(a) dy/dx = 9x2 - 5
At x = 1, dy/dx = 9(1)2 - 5 = 4
===> Gradient of tangent at x = 1 is 4
Hence, to find eqn of tangent:
y - 2 = 4 (x - 1)
y = 4x -2
(b) When dy/dx = 4, (parallel)
9x2 - 5 = 4
9x2 = 9
x2 = 1
x = 1 or -1
The other point is when x = -1
when x = -1, y = 6
Hence, the other point is (-1, 6)
