What is the equation of the normal line of f(x)=(x-1)(x+4)(x-2) at x=3 ?

1 Answer
Feb 3, 2016

23y + x - 325 = 0

Explanation:

before differentiating , distribute the brackets.

start with (x-1)(x+4) = x^2 +3x - 4

(x^2 + 3x - 4 )(x-2) = x^3-2x^2+3x^2-6x-4x+8

f(x) = x^3 + x^2 - 10x + 8

The derivative of f(x) is the gradient of the tangent and f'(3)
the value of the tangent.

f'(x) = 3x^2 + 2x - 10

and f'(3) = 3(3)^2 + 2(3) - 10 = 27+6-10 = 23 = m

For 2 perpendicular lines , the product of their gradients
is -1

let m_1 = color(black)(" gradient of normal ")

then m.m_1 = -1 → 23.m_1 = -1 rArr m_1 = -1/23

f(3) = (2)(7)(1) = 14 rArr ( 3,14) is color(black)(" normal point")

equation of normal is y-b = m(x-a) m=-1/23 , (a,b)=(3,14)

so y-14 = -1/23 (x-3)

(multiply by 23 to eliminate fraction )

23y - 322 = - x + 3 rArr 23y + x - 325 = 0