Question

Find the point on the curve y = (x²-1)(x-2) at the points where the curve cuts the X -axis .​

Answer 1

The given curve is

y = (x²-1)(x-2)

It meats at X axis so that , y = 0 at that incident .

Hence, x = +-1 and x = 2

Point is , (0,1),(0,-1),(0,2)

We want tangent ,

dy/dx = 3x²-4x-1

(dy/dx )at x = 1

dy/dx = -2

dy/dx at x = -1

dy/dx = 6

tangent at x = 2

dy/dx = 3

since, tangent at (1,0) : y - p = -2(x-1)

=> 2x+y = 2

tangent at (-1,0) : y -0 = 6(X+1)

=> y = 6x+6

tangent at (2,0)

y -0 = 3(x-2)

y = 3x-6

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Hope it's helpful !

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