# What is the equation of the normal line of #f(x)=(x-1)^2/(x-5)# at #x=4 #?

##### 1 Answer

Feb 11, 2017

#### Explanation:

So, the foot of the normal is

By actual division,

revealing asymptotes

y'=1-16/(x-5)^2=-15, at x = 4.

The slope of the normal = -1/y'=1/15.

So, the equation to the normal at

graph{((x-1)^2/(x-5)-y)(x-15y-139)((x-4)^2+(y+9)^2-1)=0 [-80, 80, -40, 40]}