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

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Answer 1 · 2015-11-10T02:08:08

#y=-4/7x+36/7#

#f'(x)=frac{d}{dx}((3x^2+2x)^{1/2})#

#=1/2(3x^2+2x)^{-1/2}(6x+2)#

The tangent line should pass through the point #(2, f(2))#, and have a gradient of #-1/{f'(2)}#.

#f(2)=4#

#f'(2)=7/4#

Equation of tangent line:

#frac{y-f(2)}{x-2}=-1/{f'(2)}#

#frac{y-4}{x-2}=-4/7#

#y=-4/7x+36/7#