What is the equation of the line normal to #f(x)=6x^3-8# at #x=7#?

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Answer 1 · 2016-12-04T01:45:25

The equation of the line normal to #f(x) = 6x^3-8# at #x=7# is:

#y-2050=-1/882(x-7)#

The equation of the tangent line to a curve #y=f(x)# at #x=barx# is:

#y-f(bar x) =f'(barx) (x -bar x)#

The normal to this line is:

#y-f(bar x) =-1/(f'(barx)) (x -bar x)#.

Calculate:

#f'(x) =18x^2#

and for #bar x =7#

#f(barx) = 2050#

#f'(bar x) = 882#

So the equation of the normal line is:

#y-2050=-1/882(x-7)#