Question

What is the equation of the normal line of `f(x)= -3x^2` at `x=0`?

Answer 1

`x=0`

Explanation:

`y = -3x^2` is a downward opening parabola, with vertex `(0,0)`.

The tangent line at the vertex is the `x`-axis and the normal line is the `y`-axis.

If you need to do this the long way

`f'(x) = -6x`

The slope of the tangent line at `x=0` is `f'(0) = 0`.

So the tangent line is a horizontal line and the normal line is the vertical line through `(0,0)`. The equation of that line is `x=0`.