Question

What is the equation of the line normal to `f(x)=3x^2 -x +1` at `x=-3`?

Answer 1

`y = 1/19 x + 592/19`

Explanation:

The function passes through the point

Denoting the point

`(x = -3, y = f(-3))`

by

`(x_1, y_1)`

The tangent to that point has instantaneous slope
`f'(-3)`

A line perpendicular to the tangent will have slope
`-1/(f'(-3))`

The line with this slope passing through `(x_1, y_1)` will be the normal to the function at that point.

Denoting its slope by `m`, t will have equation

`(y - y_1) = m (x - x_1)`

For the normal, to the function `f(x)` at `x = -3`,

`x_1 = -3`
`y_1 = f(-3)`
`m = -1/(f'(-3))`

Noting

`f(-3) = 3(-3)^2 - (-3) +1`
`= 27 + 3 + 1`
`= 31`

and
`f'(x) = 6 x - 1`

so that
`f'(-3) = 6 (-3) - 1`
`= -19`

and
`-1/(f'(-3)) = 1/19`

The equation of the normal to `f(x)` at `x = -3` is:

`(y - 31) = 1/19 (x - (-3))`

that is

`y = 1/19 x + 3/19 + 31`

that is

`y = 1/19 x + 592/19`