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

1 Answer
Nov 27, 2016

# y=2x-37/4#

Explanation:

We have # f(x) = (x-2)/(x-3)^2 #

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 # x=5 => f'(5) = (1-5)/(5-3)^3 = -4/2^3 = -1/2 #

So the gradient of the tangent when #x=5# is #m_T=1/2#

The tangent and normal are perpendicular so the product of their gradients is #-1.#

So the gradient of the normal when #x=5# is #m_N=2#

Also #f(5)=(5-2)/(5-3)^2 = 3/2^2 = 3/4#

So the normal passes through (5,3/4) and has gradient #2#, so using #y=y_1=m(x-x_1)# the required equation is;

# y-3/4=2(x-5)#
# y-3/4=2x-10#
# y=2x-37/4#

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