Show that the curve y = square root (x^2 +4x) has two slant asymptotes y = x + 2 and y = x - 2.
I am confused on where to start on this question, any help would be greatly appreciated, thanks so much in advance.
1. The second asymptote should read: y = -x - 2
2. The slope of the slant asymptote is calculated by:
$\displaystyle m = \lim_{ x \to \infty} \left(\dfrac yx \right) = \lim_{ x \to \infty} \left(\dfrac {\sqrt{x^2+4x}}x \right) = \lim_{ x \to \infty} \left(\dfrac {|x| \sqrt{1+ \frac4x}}x \right) = 1$
Or m = -1 which depends on the sign of x.
Cab you take it from here?
Nevertehless ... with your problem there exists a very elegant solution (and I have another opportunity to show off!):
$\displaystyle y=\sqrt{x^2+4x} = \sqrt{(x^2+4x+4)-4}=\sqrt{(x+2)^2-4}$
For very large values of x y is approaching
$\displaystyle a = \sqrt{(x+2)^2}=|x+2|$
... and that's the equation of the asymptote.