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

##### 1 Answer

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 #