Find the coordinates of the stationary points y=x^4-8x^2+3

Begin with the equation: y = x4-8x2+3. Differentiate by bringing the power down and reducing the power by 1 of each of the terms with x in and constant terms (3) become zero. dy/dx = 4x3-16x. Stationary point is at dy/dx = 0. 4x3-16x = 0. Solve like a normal cubic equation, x = 0, x = -2, x = 2. Sub into original equation to get y coordinate. So coordinates of stationary points are (0,3) (-2,13) and (2,13).

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Answered by Finlay H. Maths tutor

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