1. ## Slant Asymptotes

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.

2. 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?

3. I think so.

4. Originally Posted by Amybee
I think so.
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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### 24 square root 6 square root by 24 square root -6 square root

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