Find the co ordinates and nature of the turning points of the curve C withe equation, y=2x^3-5x^2-4x+2

Firstly, Differentiate the equation to find dy/dx. (dy/dx = 6x^2-10x-4)dy/dx shows the rate of change of the curve at a point, so at the stationary point dy/dx =0. So, put dy/dx = 0 and solve using factorisation. 6x^2 -10x -4=0,(x-2)(6x+2)=0x=2 or x=-1/3To find the y co ordinate plug these x values into the equation of the curve. y=2x2^3-5x2^2-4x2+2=-10y=2x(-1/3)^3-5x(-1/3)^2-4x(-1/3)+2=2.70The nature of the turning point is found using the double derrivative. So differentiate dy/dx to find d^2y/dx^2.d^2y/dx^2=12x-10where x =2:d^2y/dx^2=12x2-10=14If d^2y/dx^2>0 the point is a minimum. So, where x =2 the stationary point is a minimum.where x =-1/3:d^2y/dx^2=12x(-1/3)-10=-14If d^2y/dx^2<0 the point is a maximum. So, where x =-1/3 the stationary point is a maximum.