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QUESTION

# 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]}