The curve y = 2x^3 - ax^2 + 8x + 2 passes through the point B where x=4. Given that B is a stationary point of the curve, find the value of the constant a.
As B is a stationary point, the value of dy/dx at this point must be equal to 0. Differentiating y gives this to be dy/dx = 6x2-2ax+8. At point B, x=4. This gives the relation 104=8a and thus gives a=13.
EH
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