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

##### 1 Answer

12y + x - 109 = 0

#### Explanation:

To obtain equation of normal , y-b = m(x-a) . Require to find it's gradient m and a point on it (a , b )

differentiating f(x) will give gradient of tangent and evaluating f(x) for x = 1 will give a point on it.

differentiate using the

#color(blue)(" chain rule ") # f'(x) = 2(1 + 2x)

# d/dx (1+2x) = 2(1+2x) .2 = 4(1+2x) # and f'(1) 4(1+2) = 12 = m of tangent

now tangent and normal are perpendicular and so the product of their gradients will equal -1

If

# m_1 color(black)(" is gradient of normal") # then

# m.m_1 = -1 #

# rArr 12 xx m_1 = -1 → m_1 = -1/12 # f(1) =

# (1+2)^2 = 9 rArrcolor(black)(" (a , b ) = ( 1 , 9 ) ")# equation of normal: y- 9 =

# -1/12 (x - 1 ) #

multiplying by 12 to eliminate fraction ) gives12y - 108 = - x + 1

# rArr 12y + x -109 = 0 #