Find the turning points of the curve y=2x^3 - 3x^2 - 14.

First differentiate the equation: dy/dx = 6x^2 - 6xSet this equal to 0 as at turning points the change in gradient is 0: 0 = 6x^2 - 6x6x(x-1)=06x=0 therefore x=0(x-1)=0 therefore x=1x=1,0Now substitute back into equation 1 to find y values and hence co ordinatesx=1 y=-15x=0 y=-14

ER

Related Maths A Level answers

All answers ▸

differentiate y=(4x^3)-5/x^2


How can I find the stationary point of y = e^2x cos x?


Find the gradient of the curve y=2sinx/x^3 at the point x=


What is the binomial distribution and when should I use it?