Question

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

Answer 1

`y = -x+3`

Explanation:

First find the equation of the line TANGENT to the function at the point; then find the equation of the line that passes through that point but has a perpendicular slope.

`f(x) = x^2-x+2`

`f(1) =1^2-1+2 = 2`

`f'(x) = 2x -1`

`f'(1) = 2(1) - 1`

`f'(1) = 1`

We now have the point and the slope of the line tangent to `f(x)` at `x=1`.
`color(blue)("Tangent line: " y = 1(x-1)+2)`

Simplified tangent line: `y=x+1`
Since the slope is `1`, the slope of the normal line will be `frac{-1}{1} = -1`

`color(blue)("Normal line: " y = -1(x-1) + 2)`

Simplified normal line:
`y = -x+1+2`

`y = -x+3`