What is the equation of the normal line of #f(x)= x^3-3x^2-2x+6# at #x = 4#?

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Answer 1 · 2016-02-03T13:10:40

Equation of normal:

#y = -1/22 x + 156/11#

#f(4) = 14#

The line of normal will pass through the point #(4,14)#.

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

#f'(4) = 22#

From coordinate geometry, the gradient of the line of normal, #m#, will have a value of

#m = frac{-1}{22}#

The #y#-intercept, #c#, is given by

#c = y - mx#

#= 14 - (-1/22)*(4)#

#= 156/11#

Equation of normal:

#y = -1/22 x + 156/11#

Here is the graph

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