What is the equation of the normal line of #f(x)=-7x^3-x^2-x-1# at #x=-8#?

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Answer 1 · 2015-12-14T22:45:39

#y=1/1329x+3527#.

#f(-8)=-7(-8)^3-(-8)^2-(-8)-1=3527#.

#f'(x)=-21x^2-2x-1#.

#f'(-8)=-21(-8)^2-2(-8)-1=-1329#.

Hence the slope of the tangent at #x=-8# is #-1329#.

But the normal is perpendicular to the tangent and hence the slope of the normal at that point is #m=1/1329#.

But the normal line is a straight line so of form #y=mx+c#.

Substituting the point #(x,y)=((-8,3527)#, we get for the normal :

#3527=1/1329*(-8)+c#

#therefore c=3527#.

Thus the normal has equation #y=1/1329x+3527#.