What is the equation of the line normal to #f(x)=x^2 -x +2 # at #x=1#?

1 Answer
Aug 1, 2017

#y = -x+3#

Explanation:

First find the equation of the line TANGENT to the function at the point; then find the equation of the line that passes through that point but has a perpendicular slope.

#f(x) = x^2-x+2#

#f(1) =1^2-1+2 = 2#

#f'(x) = 2x -1#

#f'(1) = 2(1) - 1#

#f'(1) = 1#

We now have the point and the slope of the line tangent to #f(x)# at #x=1#.
#color(blue)("Tangent line: " y = 1(x-1)+2)#

Simplified tangent line: #y=x+1#
Since the slope is #1#, the slope of the normal line will be #frac{-1}{1} = -1#

#color(blue)("Normal line: " y = -1(x-1) + 2)#

Simplified normal line:
#y = -x+1+2#

#y = -x+3#

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