Question

What is the equation of the line that is normal to `f(x)=2x^3-3x^2+5` at `x=1`?

Answer 1

The normal line is the vertical line through the point `(1,4)`. Its equation is `x=1`

Explanation:

At `x=1`, we have `y=4`, so the line we want contains the point `(1,4)`

The tangent at that point has slope `f'(1)`.
Since `f'(x) = 6x^2-6x`, we get a slope of `0`.

The tangent line is horizontal. The normal line is perpendicular to the tangent line, so the normal line is vertical.