What is the equation of the line normal to #f(x)=6x^2 -x - 9 # at #x=1#?

1 Answer
Michael
Mar 10, 2016

#y=-x/11-43/11#

Explanation:

#f(x)=6x^2-x-9#

We can find the gradient of this function #m# at #x=1# by finding the 1st derivative, then we can get the gradient of the normal #m'# using the fact that #m.m'=-1#.

#f'(x)=12x-1=m#

So at #x=1:#

#m=(12xx1)-1=11#

#:.m'=-1/11#

#f(1)=6-1-9=-4#. This is the #y# value at #x=1#

The equation of the normal line is of the form:

#y=m'x+c'#

#:.-4=-1/11+c'#

#:.c'=1/11-4=1/11-44/11=-43/11#

So the equation of the normal#rArr#

#y=-x/11-43/11#

The situation looks like this:

graph{(-x/11-43/11-y)(6x^2-x-9-y)=0 [-20, 20, -10, 10]}

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