Find the equation of the tangent line to the curve y x3 X2

Find the equation of the tangent line to the curve y = x^3 - X^2 - 3 at the point on the curve where x = 1.

Solution

Given that

y = x³ - x² - 3 at the point x = 1

slope m = dy/dx

dy/dx = d/dx( x³ - x² - 3 )

= d/dx(x³) - d/dx(x²) - d/dx(3)

= 3x² - 2x - 0 [since,d/dx(xⁿ) = nx^n-1 , d/dx(constant) = 0 ]

dy/dx = 3x² - 2x

dy/dx at x = 1

dy/dx = 3(1)² - 2(1)

dy/dx = 3 - 2

= 1

Hence,

slope m = 1

Substitute x = 1 in y

y = x³ - x² - 3

= (1)³ - (1)² - 3

= 1 - 1 - 3

y = - 3

Therefore,

( x , y ) = ( 1 , -3)

let ( x₁ , y₁ ) = ( 1 , -3 )

The equation of tangent line is ,

y - y₁ = m ( x - x₁ )

y - ( -3 ) = 1. ( x - 1 )

y + 3 = x - 1

y = x - 1 - 3

y = x - 4

Therefore,

The equation of tangent is ,    y = x - 4