What is the equation of the normal line of #f(x)= -2x^3-10x # at #x=2 #?

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Answer 1 · 2017-01-12T10:53:57

The equation of the normal line is:

#y = x/34 -36-1/17#

The equation of the line normal to the graph #y=f(x)# in the point #(bar x, f(bar x))# is given by:

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

In our case #bar x = 2# and:

#f(x) = -2x^3-10x => f(2) = -36#

#f'(x) = -6x^2-10 => f'(2) = -34#

So the equation of the normal line is:

#y = -36 +1/34(x-2) = x/34 -36-1/17#