What is the equation of the curve?

Help with As Maths - equation of a curve?

• The gradient of a curve is given by dy/dx = 6x^2 - 3x The curve passes the point (-1,2). Find the equation of the curve. I have substituted in -1 into the equation and got 9. Then I put this into the equation for a curve: y-2=9(x+1) I then rearranged the equation to get y = 9x + 11. This question is worth 6 marks, so I don't think I've done enough...Can someone check my working? Thanks

• Answer:

  you'll need to integrate the dy/dx part: so 6x^2 - 3x becomes 2x^3 - 1.5x^2 + c (to integrate, increase the power by 1 and divide by the new power) now you just need to work out the + c bit, so substitute in the point they've given you... y = 2x^3 - 1.5x^2 + c 2 = (2 x -1^3) - (1.5x-1^2) + c 2 = -3.5 + c c = 5.5 so the curve equation is y = 2x^3 - 1.5x^2 + 5.5 hope this helps :)