How do you find the slant asymptote of y=sqrt(x^2+4x) y=√x2+4x?

1 Answer
George C.
Feb 24, 2016

Notice that x^2+4x = (x+2)^2 - 4x2+4x=(x+2)2−4 and take abs(x+2)|x+2| outside the square root to find two slant asymptotes:

y = x+2y=x+2

and

y = -x-2y=−x−2

Explanation:

Let f(x) = y = sqrt(x^2+4x) = sqrt(x(x+4))f(x)=y=√x2+4x=√x(x+4)

As a Real valued function, this has domain (-oo, -4] uu [0, oo)(−∞,−4]∪[0,∞), since x^2+4x >= 0x2+4x≥0 if and only if x in (-oo, -4] uu [0, oo).x∈(−∞,−4]∪[0,∞).

sqrt(x^2+4x)√x2+4x

=sqrt(x^2+4x+4-4)=√x2+4x+4−4

=sqrt((x+2)^2-4)=√(x+2)2−4

=sqrt((x+2)^2(1 - 4/((x+2)^2))= ⎷(x+2)2(1−4(x+2)2)

=abs(x+2) sqrt(1-4/(x+2)^2)=|x+2|√1−4(x+2)2

As x->+-oox→±∞ we find that 4/(x+2)^2 -> 04(x+2)2→0, so f(x)f(x) is asymptotic to abs(x+2)|x+2|

This results in two slant asymptotes:

y = x+2y=x+2 as x->+oox→+∞

and

y = -x-2y=−x−2 as x->-oox→−∞

graph{(y-sqrt(x^2+4x))(y - x - 2)(y + x + 2) = 0 [-11.01, 8.99, -1.08, 8.92]}

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