The given function
#f(x)=\frac{2-x}{\sqrt{2x+2}}#
setting #x=2# in the given function, we get #y#-coordinate of point
#y=f(2)#
#=\frac{2-2}{\sqrt{2\cdot 2+2}}=0#
Now, differentiating above equation w.r.t. #x# we get slope #dy/dx# of tangent as follows
#dy/dx=d/dxf(x)#
#=\frac{\sqrt{2x+2}(-1)-(2-x)\frac{1}{\sqrt{2x+2}}}{(\sqrt{2x+2})^2}#
#=\frac{-(2x+2)-(2-x)}{(2x+2)^{3/2}}#
#=\frac{-x-4}{(2x+2)^{3/2}}#
setting #x=2# in above equation, we get slope of tangent at #(2, 0)#
#=\frac{-2-4}{(2\cdot 2+2)^{3/2}}#
#=-1/\sqrt6#
hence the slope #m# of normal at the same point
#=-1/(-1/\sqrt6)=\sqrt6#
hence the equation of normal at #(x_1, y_1)\equiv (2, 0) # & having slope #m=\sqrt6# is given by following formula
#y-y_1=m(x-x_1)#
#y-0=\sqrt6(x-2)#
#y=\sqrt6(x-2)#
