Question 12: Given the function \(y = \frac{1}{4}{x^4} – 2{x^2} + 3\) whose graph is (C). The area of the triangle whose vertices are the extreme points of the graph (C) is:

We have:\(y = \frac{1}{4}{x^4} – 2{x^2} + 3\)

Extreme points: 2;−first); B(0,3); C(2;−1)

The extreme points form an isosceles triangle at B and H(0;−1) is the midpoint of AC

So \({S_{\Delta ABC}} = \frac{1}{2}BH.AC = 8\)

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