Question

What is the equation of the normal line of `f(x)= (x^2-x)/(x-3)` at `x = 4`?

Answer 1

I found: `y=1/5x+56/5`

Explanation:

First we evaluate the derivative of the function at `x=4`:
`f'(x)=((2x-1)(x-3)-(x^2-x))/(x-3)^2` using the Quotient Rule.
Evaluate it at `x=4`:
`f'(4)=(((7*1)-12))/1=-5=m`

This is the slope of the TANGENT to our curve at `x=4`; to get the NORMAL we use `m'=-1/m=1/5` as slope.

The point where our normal passes on the curve represented by our function, has coordinates:
`x_0=4`
`f(4)=y_0=(16-4)/1=12`
and the equation of the line through this point with slope `m'` is given as:
`(y-y_0)=m'(x-x_0)`
`y-12=1/5(x-4)`
`y=1/5x+56/5`