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

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

(x³) – (2x²) – (2x)

⇒ = 3x^{3 – 1} – (x^{2 – 1}) – 2x^{1 – 1}

⇒ = 3x² – 4x – 2 ...(1)

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

so the 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 Quadratic equation.

⇒ 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 =

Substitute x = 2 & x = in y = x³ – 2x² – 2x

when x = 2

⇒ y = (2)³ – 2 (2)² – 2(2)

⇒ y = 8 – (24) – 4

⇒ y = 8 – 8 – 4

⇒ y = – 4

when x =

⇒ y = ()³ – 2 ()² – 2()

⇒ y = () – 2 () + ()

⇒ y = () – () + ()

taking lcm

⇒ y =

⇒ y =

⇒ y =

Thus, the points are (2, – 4) & (,)