# What is the equation of the line that is normal to #f(x)= x(x-4) # at # x= 5 #?

#f(x) = x^2-4x

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The equation of the line that is normal to f(x) = x(x-4) at x = 5 is y = -1/6(x-5) + 5.

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The equation of the line that is normal to ( f(x) = x(x-4) ) at ( x = 5 ) is ( y = -\frac{1}{f'(5)}(x - 5) + f(5) ). First, find the derivative of ( f(x) ), then evaluate it at ( x = 5 ) to get the slope of the tangent line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. Finally, use the point-slope form of a line with the given point ( (5, f(5)) ) and the slope calculated to find the equation of the normal line.

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