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

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Answer 1 · 2016-11-07T00:31:41

$x=0$

$y = -3x^2$ 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 $x=0$ .

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