What is the equation of the normal line of #f(x)=-x^4+2x^3-x^2-9x-3# at #x=-1#?

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Answer 1 · 2015-11-24T02:12:17

#y = -1/3x+5/3#

#f(-1) = -1-2-1+9-3 =2#

#f'(x) = -4x^3+6x^2-2x-9#

so #f'(-1) = 4+6+2-9 = 3#.

The slope of the tangent line at #(-1,2)# is #3#, so the normal line has slope #-1/3#.

The line with slope #m=-1/3# through the point #(-1,2)# is

#y-2 = -1/3(x-(-1))#

#y-2 = -1/3x-1/3#

#y = -1/3x+5/3#