Question

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

Answer 1

`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`