# What is the equation of the normal line of #f(x)=(x-2)/(x-3)^2# at #x=5#?

##### 1 Answer

# y=2x-37/4#

#### Explanation:

We have

Using the quotient rule we have:

# f'(x) = { (x-3)^2 d/dx(x-2) - (x-2)d/dx(x-3)^2 }/ ((x-3)^2)^2#

# :. f'(x) = { (x-3)^2 (1) - (x-2)2(x-3)(1) }/ ((x-3)^4)#

# :. f'(x) = { (x-3){ (x-3) - 2(x-2)} }/ ((x-3)^4)#

# :. f'(x) = { (x-3 - 2x+4) }/ ((x-3)^3)#

# :. f'(x) =(1-x)/ ((x-3)^3)#

When

So the gradient of the tangent when

The tangent and normal are perpendicular so the product of their gradients is

So the gradient of the normal when

Also

So the normal passes through (5,3/4) and has gradient

# y-3/4=2(x-5)#

# y-3/4=2x-10#

# y=2x-37/4#