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

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Answer 1 · 2018-01-08T18:40:23

#y = 1/2x#

I'll start by finding the slope of the line normal to #f(x)# 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)

#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#