What is the equation of the normal line of #f(x)=5x^3-2x^2-3x-1# at #x=-8#?

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Answer 1 · 2017-07-30T06:27:55

Equation of normal is #x+989y+2635693=0#

At #x=-8#, #f(x)=5(-8)^3-2(-8)^2-3(-8)-1#

#=-2560-128+24-1=-2665#

Hencewe are seeking normal at #(-8,-2665)#

Slope of tangent is given by #f'(-8)# and as #f'(x)=15x^2-4x-3#

slope of tangent is #15(-8)^2-4(-8)-3=989#

and hence slope of normal is #-1/989# and

equation of normal is #y+2665=-1/989(x+8)#

or #x+989y+2635693=0#