Find the equation of the line tangent to fx x2 2x 8 at the

Find the equation of the line tangent to f(x) = x^2 + 2x -8 at the point where x = 1.

Solution

Given that

Let y = f(x) = x² + 2x - 8 at point x = 1

slope m = f\'(x)

dy/dx = d/dx( x² + 2x - 8 )

= d/dx(x²) + 2.d/dx(x) - d/dx(8)

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

f\'(x) = 2x + 2

f\'(x) at x = 1

f\'(x) = 2x + 2

f\'(1) = 2.1 + 2

= 2 + 2

= 4

Hence,

slope m = 4

Substitute x = 1 in f(x)

y = x² + 2x - 8

= (1)² + 2(1) - 8

= 3 - 8

y = -5

Therefore,

( x , y ) = ( 1 ,- 5 )

let ( x₁ , y₁ ) = ( 1 ,- 5 )

The equation of tangent line is ,

y - y₁ = m ( x - x₁ )

y - ( -5) = 4 ( x - 1 )

y + 5 = 4x - 4

y = 4x - 4 - 5

y = 4x - 9

Therefore,

The equation of tangent line is ,    y = 4x - 9