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

Feb 11, 2017

$x - 15 y - 139 = 0$. See the normal-inclusive Socratic graph.

#### Explanation:

$y = - 9$, at $x = 4$.

So, the foot of the normal is $P \left(4 , - 9\right)$.

By actual division,

$y = x + 3 + \frac{16}{x - 5}$,

revealing asymptotes $y = x + 3 \mathmr{and} x = 5$.

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 $P \left(4 , - 9\right)$ is

$y + 9 = 15 \left(x - 4\right)$, giving

$x - 15 y - 139 = 0$

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