How do you find the slant asymptote of y=sqrt(x^2+4x) ?
1 Answer
Feb 24, 2016
Notice that
y = x+2
and
y = -x-2
Explanation:
Let
As a Real valued function, this has domain
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
This results in two slant asymptotes:
y = x+2 asx->+oo
and
y = -x-2 asx->-oo
graph{(y-sqrt(x^2+4x))(y - x - 2)(y + x + 2) = 0 [-11.01, 8.99, -1.08, 8.92]}