What is the equation of the normal line of f(x)= (x^2-x)/(x-3)f(x)=x2xx3 at x = 4x=4?

1 Answer
Nov 11, 2015

I found: y=1/5x+56/5y=15x+565

Explanation:

First we evaluate the derivative of the function at x=4x=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