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

1 Answer
Nov 11, 2015

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#