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]}