# What is the equation of the line normal to #f(x)= -1/(3-2x) # at #x=0#?

##### 1 Answer

Jan 17, 2016

# y = -9/2 x + 1/3 #

#### Explanation:

rewrite f(x) as f(x) =

# (3 - 2x )^-1 # ( using the 'chain rule' ) gives:

# f'(x) = - 1 (3 - 2x )^-2 . d/dx ( 3 - 2x ) #

#rArr f'(x) = - (3 - 2x )^-2 (-2) = 2/(3 - 2x )^2 # at x = 0 :

# f'(0) = 2/3^2 = 2/9# and f(0) = 1/3

For perpendicular lines

# m_1.m_2 = - 1 rArr m_2 = (-1)/(2/9) = -9/2 # equation of normal : y - b = m (x - a )

where

# m = -9/2 ( a , b ) = ( 0 , 1/3 )# ( substitute in values ) :

# y - 1/3 = -9/2 x #

#rArr y = -9/2 x + 1/3 #