Question

What is the equation of the normal line of `f(x)=x^2-x+5` at `x=2`?

Answer 1

`y=(23-x)/3`

Explanation:

Differentiate:
`f prime(x)=2x-1`
`f prime(2)= 2(2)-1`
`=3`
Therefore the normal has gradient `-1/3` because "`m_1m_2=-1`"for perpendicular lines.
Therefore the equation of the normal line is `y=-x/3+c`.
This has to pass through the point `(1,f(2))`, that is, `(1,7)`.
Therefore `7=-2/3+c` hence `c=23/3`.
So the line is `y=-x/3+23/3`.