What is the equation of the line normal to # f(x)=-sqrt((x+1)(x+3)# at # x=0#?

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Answer 1 · 2018-07-30T17:33:30

#y=sqrt(3)/2x-sqrt(3)#

Given #f(x)=-sqrt((x+1)(x+3))#
we get #f(0)=-sqrt(3)#

Now we will compute the slope of the normal line:

#f'(x)=-((x+1)(x+3))^(-1/2)*(x+3+x+1)#so
#f'(x)=-(4+2x)/(2sqrt((x+1)(x+3))#

#f'(0)=-2/sqrt(3)# so the slope of our normal line is given by

#m_N=sqrt(3)/2#

#y=sqrt(3)/2x+n#

The Point is given by #(0,-sqrt(3))#

and we get #n=-sqrt(3)#

therefore our line is given by

#y=sqrt(3)/2x-sqrt(3)#