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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` |
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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)` Answer |
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