What is the equation of the line normal to #f(x)=3x^2 - 12x-5 # at #x=1#?

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Answer 1 · 2016-01-03T15:17:56

So the equation normal to #f(x)# at #x=1# is

#y=1/6x-85/6#

The normal is a line at right angles to the tangent.

First we need to calculate the #df(1)/dx# hence we have that

#f'(x)=6x-12=>f'(1)==-6#

So the normal will have a gradient of #m*(-6)=-1=>m=1/6#

hence at point #(1,f(1))# or #(1,-14)# the equation is

#y=m*x+k=>y=1/6*x+k#

replacing the value of point #(1,f(1))# we get

#-14=1/6+k=>k=-14-1/6=>k=-85/6#