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

Question

13759 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-24T21:51:53

Notice that #x^2+4x = (x+2)^2 - 4# and take #abs(x+2)# outside the square root to find two slant asymptotes:

#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 domain #(-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]}