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Question -

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



Answer -

Given the curve y = x3 – 2x2 –2x and a line y = 2x – 3

First, we will find the slope of tangent

y = x3 – 2x2 – 2x

y = 2x – 3 is the form of equation of a straight liney = mx + c, where m is the Slope of the line.

So the slope of the line is y = 2 × (x) – 3

Thus, the Slope = 2. … (2)

From (1) & (2)

 3x2 –4x – 2 = 2

 3x2 –4x = 4

 3x2 –4x – 4 = 0

We will use factorization method to solve the above Quadraticequation.

 3x2 –6x + 2x – 4 = 0

 3x (x – 2) + 2 (x – 2) = 0

 (x– 2) (3x + 2) = 0

 (x– 2) = 0 & (3x + 2) = 0

 x= 2 or

x = -2/3

Substitute x = 2 & x = -2/3 in y = x3 –2x2 – 2x

When x = 2

 y= (2)3 – 2 × (2)2 – 2 × (2)

 y= 8 – (2 × 4) – 4

 y= 8 – 8 – 4

 y= – 4

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