What is the equation of the line normal to # f(x)=2/(x-1)^2-2x+4# at # x=-2#?

Question

1235 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-20T12:54:47

Equation of normal is #y = 27/50 x+ 2093/225#

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

#f(-2) =2/(-2-1)^2-2 *(-2) +4 = 2/9+4+4=74/9 #

The point is #(-2, 74/9)# at which normal to be drawn.

#f^'(x) =-4/(x-1)^3-2 # . Slope of tangent at #x=-2# is

#f^'(-2) =-4/(-2-1)^3-2 = 4/27-2= -50/27 #

Slope of normal at #x=-2# is #m= 27/50#

Equation of normal at point #(-2, 74/9)# is

#y - 74/9 = 27/50 (x+2)# or

#y = 27/50 x+ 27/25 +74/9# or

#y = 27/50 x+ 2093/225# [Ans]