Find the points on the curve `y=x^3-2x^2-x` at which the tangent lines are parallel to the line `y=3x-2` .

Explanation

`y=mx+c,`slope of this line=m

`:. y=3x-2`, slope of line `y=3x-2 `is 3

slope of tangent of any curve is `dy/dx` at that point

`y= x^3 -2x^2 - x`

`dy/dx = d/dx(x^3 - 2x^2 - x) `

`= 3x^2 - 4x-1`

`:. dy/dx = 3`

`=> 3x^2 - 4x - 1 = 3`

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