To draw a tangent at a point #(x_0,f(x_0))# on the curve #f(x)#, we need to find the slope of the tangent, which is given by value of #f'(x)=(df)/(dx)# at that point i.e. slope is #f'(x_0)# and as it passes through #(x_0,f(x_0))#, its equation is
#y-f(x_0)=f'(x_0)(x-x_0)#
The slope of normal is #-1/(f'(x_0))# and its equation is
#y-f(x_0)=-1/(f'(x_0))(x-x_0)#
Here we are seeking tangent to function #f(x)=x^2/(x+2)# at #(2,1)# and note that #f(2)=2^2/(2+2)=1#
Now using quotient formula #f'(x)=(2x(x+2)-1(x^2))/(x+2)^2#
= #(2x^2+4x-x^2)/(x+2)^2=(x^2+4x)/(x+2)^2#
and as #f'(2)=(2^2+4*2)/(2+2)^2=12/16=3/4#, slope of tangent is #3/4# and slope of normal is #-1/(3/4)=-4/3#
Hence equation of tangent is
#y-1=3/4(x-2)# or #4y-4=3x-6# i.e. #3x-4y-2=0#
and slope of normal is
#y-1=-4/3(x-2)# or #3y-3=-4x+8# i.e. #4x+3y-11=0#
graph{(xy+2y-x^2)((x-2)^2+(y-1)^2-0.02)(4x+3y-11)(3x-4y-2)=0 [-10, 10, -5, 5]}
