Given that y=x^3 +2x^2, find dy/dx . Hence find the x-coordinates of the two points on the curve where the gradient is 4.

First of all we are asked to differentiate the function. This can be easily done by multiplying the coefficient of x by its exponent, and then decreasing the exponent by one. Therefore, dy/dx=3x^2+4x We are then asked to find the two points on the curve where the gradient is 4; this can be solved by setting the derivative we just calculated to equal four ( since the derivative function gives us the value of the gradient at each x-coordinate). Which gives us: 3x^2+4x=4 = 3x^2+4x-4=0 we can solve this quadratic by using the quadratic formula, which gives us the solutions: x=-2 and x=2/3