Find the turning points on the curve with the equation y=x^4-12x^2

y = x^4 - 12x^2
dy/dx = 4x^3 - 24x
The turning points are where dy/dx = 0
4x^3 - 24x =0
x(4x^2 - 24) = 0 Therefore one of the turning points is at x = 0
4x^2 - 24 = 0
4x^2 = 24
x^2 = 6
x = +/- √6
Substitute the x coordinates back into the original equation to find y
The final coordinates are (0,0), (√6,-36) and (-√6,-36)

Related Maths A Level answers

All answers ▸

integrate from 0 to 2: 2x*sqrt(x+2) dx


If given two parametric equations for a curve, how would you work out an equation for the gradient?


Solve for x (where 0<x<360) 2sin^2(x) - sin(x) - 1 = 0


Find the x-coordinates of any stationary points of the equation y = x^3 - 2x + 4/x