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

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Answer 1 · 2017-02-11T09:21:43

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

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

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

By actual division,

$y = x+3+16/(x-5)$ ,

revealing asymptotes $y=x+3 and x =5$ .

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 $P( 4, -9 )$ is

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

$x-15y-139=0$

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

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