Application of Derivatives

Find the point on the curve y = x 3 – 2x 2 – 2x at which the tangent lines are parallel to the line y = 2x – 3. - Wired Faculty

Question

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Find the point on the curve y = x³ – 2x² – 2x at which the tangent lines are parallel to the line y = 2x – 3.

Solution

The equation of curve is y = x³ – 2x² – 2x
$therefore space space space dy over dx space equals space 3 straight x squared minus 4 straight x minus 2 therefore space space space space space at space left parenthesis straight x comma space space straight y right parenthesis comma space space slope space of space tangent space space equals space 3 straight x squared minus 4 straight x minus 2 therefore space space space space space space straight m subscript 1 space equals space 3 straight x squared minus 4 straight x minus 2 comma space space where space straight m subscript 1 space is space slope space of space tangent Let space space straight m subscript 2 space be space slope space of space line space straight y space equals space 2 straight x minus 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis therefore space space space space space space space straight m subscript 2 space equals space 2 because space space space space space space space space space tangent space is space parallel space to space line space left parenthesis 2 right parenthesis therefore space space space space space space space space space space space space space space straight m subscript 1 space equals space straight m subscript 2 rightwards double arrow space space space space space space straight x space equals space fraction numerator 4 plus-or-minus square root of 16 plus 48 end root over denominator 6 end fraction space equals space fraction numerator 4 plus-or-minus 8 over denominator 6 end fraction space equals space 2 comma space space space minus 2 over 3 When space straight x space equals space 2 comma space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space 8 minus 8 space minus 4 space equals space minus 4 When space straight x space equals space minus 2 over 3 comma space space from space left parenthesis 1 right parenthesis comma space space space straight y space equals space minus 8 over 27 minus 8 over 9 plus 4 over 3 space equals space fraction numerator negative 8 minus 24 plus 36 over denominator 27 end fraction space equals space 4 over 27 therefore space space space space space required space points space are space left parenthesis 2 comma space minus 4 right parenthesis comma space open parentheses negative 2 over 3 comma space space space space 4 over 27 close parentheses$

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