At any point on a curve d2y/dx^2 = 6x - 2. Given that the tangent to the curve at (1,4) is y = 2x + 2, find the equation of the curve.
Hmm, quite a tough question. Or so I think. Please help. thanks.
d2y/dx2 = 6x-2 ==> dy/dx=3x^2 - 2x + c (where c is a constant)
you know from the qn that the tangent to the curve at (1,4) is y = 2x + 2, so it means that dy/dx at x=1 has a gradient of 2, so... 3(1)^2 - 2(1) + c = 2 ==> c = 1
therefore dy/dx = 3x^2 - 2x + 1
then you integrate again to find that it has another constant d, but it also contains the point (1,4). i think you can continue on yourself =)