What is the equation of the normal line of #f(x)= 3x - 2x^2# at #x=1#?

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Answer 1 · 2016-01-15T10:02:31

#y=x#

Given: #y=3x-2x^2#

Computing for the ordinate #y=1#

our point is #(1, 1)#

#f' (x)# = slope

#f' (1)=-1#

the negative reciprocal of the slope is needed foe the Normal line.

so for the normal line #m=-1/(f' (1))=-1/-1=1#

#y-y_1=m(x-x_1)#