The tangent equation of curve y = x3-2x2-4x + 2 at points (1, - 3) is () A. 5x+y+2=0B. 5x+y-2=0C. 5x-y-8=0D. 5x-y+8=0
∵ curve y = x3-2x2-4x + 2, ∵ y ′ = 3x2-4x-4, when x = 1, y ′ = - 5, that is, tangent slope is - 5, ∵ tangent equation is y + 3 = - 5 (x-1), that is, 5x + Y-2 = 0
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