Question

What is the equation of the line normal to `f(x)=x^3-6x^2-2` at `x=1`?

Answer 1

`y+7=1/9(x-1)`

Explanation:

Find the slope of the tangent line, to which the normal line is perpendicular.

`f(1)=-7`

The line will pass through the point `(1,-7)`.

`f'(x)=3x^2-12x`

`f'(1)=-9`

If the slope of the tangent line is `-9`, the slope of the normal line will be the opposite reciprocal of `-9` since the lines are perpendicular. This means the slope is `1/9`.

Write the equation in point-slope form:

`y+7=1/9(x-1)`