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

Answer 1

Aurora Alvarado

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#

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Answer 2

Samantha Adkison

The equation of the normal line of f(x)=5x^3-2x^2-3x-1 at x=-8 is y = -120x - 319.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.