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 -1let
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