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

Nov 7, 2016

$x = 0$

#### Explanation:

$y = - 3 {x}^{2}$ is a downward opening parabola, with vertex $\left(0 , 0\right)$.

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

$f ' \left(x\right) = - 6 x$

The slope of the tangent line at $x = 0$ is $f ' \left(0\right) = 0$.

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