What is the equation of the normal line of #f(x)=(x+2)/(x-4)# at #x=3#?

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Answer 1 · 2016-04-05T05:52:51

Equation of the normal line is #x-2y-13=0#

At #x=3#, #f(3)=(3+2)/(3-4)=5/-1=-5#

Hence normal is desired at point #(3,-5)# on the curve.

Slope of tangent is given by #f'(x)#

= #((x-4)*d/dx(x+2)-(x+2)*d/dx(x-4))/(x-4)^2#

= #((x-4)*1-(x+2)*1)/(x-4)^2=(x-4-x-2)/(x-4)^2=-2/(x-4)^2#

Hence slope of tangent at #(3,-5)# is #-2/(3-4)^2=-2#.

Hence slope of normal would be #-1/-2=1/2#

And equation of the normal line using point slope form is

#(y-(-5))=1/2(x-3)# or

#2*(y+5)=(x-3)# or #x-2y-13=0#