Integration Help

The curve C with the equation y=f(x) is such that,
dy/dx = 3x^2 -4x - 1
Given that the tangent to the curve at the point P with x coord 2 passes through the origin, find an equation for the curve.

Ive integrated it and got x^3 - 2x^2 - x + c
Im not sure how to get c ( which is +8)

Any help is much appreciated

Stick 2 into dy/dx to give the gradient of the tangent, which gives you 3. Since the tangent passes through the origin, equation for the tangent must be y = 3x. Substituting 2 into y = 3x gives us y = 6 and so point is P is (2,6). Now substitute 2 and 6 into your integrated form and rearrange to find that c = 8.

Key point here is "the tangent to the curve at Point P with x-coordinate 2 passes through the origin"

The tangent passes through the origin, so the equation for the tangent is of the form y = mx. That lets you find the y-coordinate of point P and hence the constant term in the function.

You can find the gradient of the tangent by substituting x=2 into the derivative of curve C. You can then find the equation of the tangent using y-y₁=m(x-x₁) and substituting point (0,0) to get y=3x. You can then find the point that this tangent touches the curve by substituting x=2 into y=3x, to get point (2,6). Then substitute y=6 and x=2 into the expression for the curve, and you should get 6= c - 2. Then solve for c to get c=8.