Question

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

Answer 1

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`