Question

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

Answer 1

`y=x-2`

Explanation:

first find `f'(x)`
`f(x)=(x-2)^(-1)`
`f'(x)=-(x-2)^(-2)`

then find the slope of the tangent line at `x=1`, which is also `f'(1)`
`f'(1)=-(1-2)^(-2)`
`f'(1)=-(-1)^(-2)`
`f'(1)=-1`

use this fact: normal slope `*` tangent slope `= -1`

normal slope `*(-1)=-1`
normal slope `=1`

now find the point where `x=1`
the y-value is `f(1)=1/(1-2)=1/(-1)=-1`

for the normal line, the point is `(1,-1)` and the slope is 1

use point-slope form:
`y-(-1)=1(x-1)`
`y+1=x-1`
`y=x-2`