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

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Answer 1 · 2016-01-09T04:43:49

#y=-x/6-11/3# or #x+6y=-22#

the given: #f(x) = x^3-6x# at #x=2#

Find the value of the function at #x=2#

#f(2) = (2)^3-6(2)=8-12=-4#

#f(2) = -4#

The point on the curve is #(2, -4)#

Find first derivative #f' (x) # then evaluate at #x=2#

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

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

#m=6# to be used for Tangent line.

#m#(perpendicular)#=-1/6# to be used for the Normal Line

Use now the #m=-1/6# and the point #(2,-4)#

The Normal Line:

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

#y-(-4)=(-1/6)(x-2)#

#y+4=(-1/6)(x-2)#

#6y+24=-x+2#

#6y=-x-22#

#y=-x/6-11/3# the required Normal Line