What is the equation of the normal line of #f(x)=x^3-8x^2+9x-3# at #x=-1#?

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Answer 1 · 2016-11-22T08:07:45

#f(-1)=-21#

#f'(x) = 3x^2-16x+9#

#f'(-1) = 3+16+9=28 = m_tan#

The slope of the normal line at #x=-1# is the negative reciprocal of the slope of the tangent line #m=28# which would be #-1/28#.

So, using a slope formula:

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

We get:

#y+21=-1/28(x+1)#

#y=-1/28x-589/28#