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

Jan 8, 2018

$y = \frac{1}{2} x$

#### Explanation:

I'll start by finding the slope of the line normal to $f \left(x\right)$ at $x = 0$.

The line normal to f(x) will be perpendicular to the line tangent to f(x), which will have a slope of f'(x).

Therefore, the slope of the line normal to f(x) will be 1/f'(x)

$\frac{1}{f ' \left(x\right)} = \frac{1}{2 - 2 x}$

Based on this the slope of the line normal to f(x) at x = 0, will be
$\frac{1}{2 - 2 \cdot 0} = \frac{1}{2}$

We know that f(x) passes through the point $\left(0 , f \left(0\right)\right) = \left(0 , 0\right)$ at x = 0, so our line normal to f(x) at x = 0 is:

$y = \frac{1}{2} x$