find the coordinates of the turning points of the curve y = 2x^4-4x^3+3, and determine the nature of these points

To begin, we must first use the fact that turning points of a graph occur at points where the gradient is equal to zero, in other words, points where dy/dx =0. Differentiating the equation with and setting to zero gives dy/dx = 8x³-12x²=0, then, solving for x we get x = 0 and x = 3/2. Putting these x values back into the original equation will give us the coordinates of the turning points which are (0,3) and(3/2,-3/8).The second part of the question asks us to determine the nature of the turning points, for which we will have to use the second derivative. Differentiating dy/dx again gives d²y/dx²=24x²-24x. at x= 3/2, d²y/dx²= 18 which makes it a minimum point since d²y/dx²>0. x=0, d²y/dx²=0 which means it could either be a minimum, maximum, or point of inflection, we will have to run further tests to determine the nature of this point.