What is the equation of the normal line of #f(x)=-2x^3+18x^2-25x# at #x=-1/3#?

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Answer 1 · 2017-10-30T15:15:52

#y=-3/113*x+31726/3051#

I took diferential both sides,

#f'(x)=-6x^2+36x-25#

#f'(-1/3)=-2/3-12-25=-113/3#

Hence I found slope of normal line: #-3/113#

I found a point, which passes from it:

#f(-1/3)=2/27+18/9+25/3=281/27#

Thus,

#y-281/27=-3/113*(x+1/3)#

#y-281/27=-3/113*x-1/113#

#y=-3/113*x+31726/3051#