Question

Find the points on the curve y = x³ - 2x² - x, where the tangents are parallel to 3x - y + 1 = 0

Answer 1

Hey Mate
your answer is here

Slope of given line = 3

and

dy/dx = 3x^2-4x-1

so,

3 = 3x^2-4x-1

3x^2-4x-4 = 0

so,

X = 2 & -2/3



Thanks

Answer 2

Answer:

(2,-2) and (-2/3 , -14/27)

Step-by-step explanation:

Given in the question a curve y = x³ - 2x² - x and an equation of straight line

3x - y + 1 = 0

Step 1

Since the Tangent is parallel to the line 3x - y + 1 = 0, so there gradient will be same

To find gradient of this line we will arrange it in the standard form of equation

y = mx +c

here m is the gradient

3x - y + 1 = 0

- y = -1 -3x

 y = 3x + 1

so,

m = 3

Step 2

Take derivative of the curve equation

dy/dx = x³ - 2x² - x

         = 3x² - 4x - 1

Step3

Equal the derivative to the gradient found and solve the equation

3 = 3x² - 4x - 1

3x² - 4x - 1 - 3 = 0

3x² - 4x - 4 = 0

(x-2)(x+2/3) =0

x1 = 2

x2 = -2/3

Step 4

Equate the value of x in the curve equation

y = 2³ - 2(2)² - (2)

y = -2

y = -2/3³ - 2(-2/3)² - 2/3

y = -14/27

Step 5

points on the curve  are

(2,-2) and (-2/3,-14/27)