#f(x) = x^2/(x+5) = ((x^2 + 5x) - (5x + 25) + 25)/(x+5)#
#=((x-5)(x+5)+25)/(x+5)#
#=x-5 + 25/(x+5)#
For large positive or negative #x# this will be asymptotic to #x - 5#
#f(x)# has a simple pole at #x = -5# with #f(-5-epsilon)# being large and negative, and #f(-5+epsilon)# is large and positive for small #epsilon > 0#
#f(0) = 0# so the curve passes through #(0, 0)#
#f'(x) = (2x)/(x+5)-x^2/(x+5)^2#
#=((2x)(x+5)-x^2)/(x+5)^2#
#=(x^2+10x)/(x+5)^2#
#=(x(x+10))/(x+5)^2#
So #f'(x) = 0# when #x = 0# and #x = -10#
So there is a local minimum at #(0, 0)# and a local maximum at #(-10, f(-10)) = (-10, -20)#
graph{x^2/(x+5) [-86.4, 73.6, -45.1, 34.9]}
