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

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Answer 1 · 2018-01-08T18:40:23

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

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

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)

$1/{f'(x)}=1/{2-2x}$

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

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

$y = 1/2x$

What is the equation of the line normal to f(x)=2x-x^2 f(x)=2x−x2f(x)=2x-x^2 at x=0x=0x=0?