1.

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`

Answer» `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`
`=> 3x^2 + 2x - 6x - 4= 0`
`:. x(3x+2) - 2(3x+2) => (x-2)(3x+2) = 0`
`:. x=2 or -2/3`
for x=2`=> y= (2)^3 - 2(2)^2 - 2 = 8-8-2= -2`
for x=-2/3 `=> y= (-2/3)^3 - 2(2/3)^2- (-2/3) `
`= -8/27 - 8/9 + 2/3`
`= (-8+24+18)/27 = -14/27`
`:. ` points are `(2,-2); (-2/3.-14/27)`
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