Slant Asymptotes

**Amybee** · November 22nd, 2009, 08:36 PM

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.

**earboth** · November 22nd, 2009, 10:03 PM

Originally Posted by Amybee

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:

$m = \lim_{ x \to \infty} \left(\dfrac yx \right) = \lim_{ x \to \infty} \left(\dfrac {\sqrt{x^2+4x}}x \right) = \lim_{ x \to \infty} \left(\dfrac {|x| \sqrt{1+ \frac4x}}x \right) = 1$
Or m = -1 which depends on the sign of x.

Cab you take it from here?

**Amybee** · November 22nd, 2009, 10:20 PM

I think so.
Thanks for your help!

**earboth** · November 23rd, 2009, 05:34 AM

Originally Posted by Amybee

I think so.
Thanks for your help!

Nevertehless ... with your problem there exists a very elegant solution (and I have another opportunity to show off!):

$y=\sqrt{x^2+4x} = \sqrt{(x^2+4x+4)-4}=\sqrt{(x+2)^2-4}$

For very large values of x y is approaching

$a = \sqrt{(x+2)^2}=|x+2|$

... and that's the equation of the asymptote.

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