Equation of normal at #x=x_0# on the curve #y=f(x)# is perpendicular to the tangent at #x=x_0# at #f(x)#.
As slope of tangent at #x=x_0# on the curve #f(x)#, is given by #f'(x_0)#, slope of normal is #-1/(f(x_0))# and equation of normal is
#y=-1/(f(x_0))(x-x_0)+f(x_0)#
Here, we have #y=f(x)=-sqrt((x+1)(x+3))#
and #f'(x)=(df)/(dx)# - using chain rule
= #-1xx1/(2sqrt((x+1)(x+3)))[1xx(x+3)+1xx(x+1)]#
= #-1/(2sqrt((x+1)(x+3)))(2x+4)=-(x+2)/sqrt((x+1)(x+3))#
and while #f(0)=-sqrt3#, #f'(0)=-2/sqrt3#
as such slope of normal at #x=0# is #sqrt3/2# and
equation of normal is #y=sqrt3/2x-sqrt3#
graph{(y-sqrt3/2x+sqrt3)(y+sqrt((x+1)(x+3)))=0 [-12.92, 7.08, -8.52, 1.48]}
The curve or normal is not appearing in the interval #(-1,-3)# as #f(x)=-sqrt((x+1)(x+3))# is not defined in the interval #(-1,-3)#.
