How do you find the slant asymptote of y=sqrt(x^2+4x) ?

Notice that ##x^2+4x = (x+2)^2 – 4## and take ##abs(x+2)## outside the square root to find two slant :
##y = x+2##
and
##y = -x-2##
Let ##f(x) = y = sqrt(x^2+4x) = sqrt(x(x+4))##
As a Real valued function this has ##(-oo -4] uu [0 oo)## since ##x^2+4x >= 0## if and only if ##x in (-oo -4] uu [0 oo).##
##sqrt(x^2+4x)##
##=sqrt(x^2+4x+4-4)##
##=sqrt((x+2)^2-4)##
##=sqrt((x+2)^2(1 – 4/((x+2)^2))##
##=abs(x+2) sqrt(1-4/(x+2)^2)##
As ##x->+-oo## we find that ##4/(x+2)^2 -> 0## so ##f(x)## is asymptotic to ##abs(x+2)##
This results in two slant asymptotes:
##y = x+2## as ##x->+oo##
and
##y = -x-2## as ##x->-oo##
graph{(y-sqrt(x^2+4x))(y – x – 2)(y + x + 2) = 0 [-11.01 8.99 -1.08 8.92]}