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

Apr 4, 2018

$y = x - 2$

#### Explanation:

first find $f ' \left(x\right)$
$f \left(x\right) = {\left(x - 2\right)}^{- 1}$
$f ' \left(x\right) = - {\left(x - 2\right)}^{- 2}$

then find the slope of the tangent line at $x = 1$, which is also $f ' \left(1\right)$
$f ' \left(1\right) = - {\left(1 - 2\right)}^{- 2}$
$f ' \left(1\right) = - {\left(- 1\right)}^{- 2}$
$f ' \left(1\right) = - 1$

use this fact: normal slope $\cdot$ tangent slope $= - 1$

normal slope $\cdot \left(- 1\right) = - 1$
normal slope $= 1$

now find the point where $x = 1$
the y-value is $f \left(1\right) = \frac{1}{1 - 2} = \frac{1}{- 1} = - 1$

for the normal line, the point is $\left(1 , - 1\right)$ and the slope is 1

use point-slope form:
$y - \left(- 1\right) = 1 \left(x - 1\right)$
$y + 1 = x - 1$
$y = x - 2$