show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
Slant asymptotes are lines that the curve approaches as x --> +oo or x --> -oo. As x --> +oo or -oo, and the square root of this is . This is not a rigorous way of doing it. The mathematicians of MHF will pounce on this and make mince meat of it.
But I'm not sure you're at a level where a rigorous approach would make much sense, so you'll just have to make do with the one above.
Hi all. I was confused by this limit too. I am not quite sure what is wrong with this reasoning. I have a suspicion but I would like to see other thoughts.
For x > 0, sqrt(x^2 + 4x) is the same as (x^2 + 4x)/sqrt(x^2 + 4x) = (x + 4)/sqrt(1 + 4/x).
At this point, why can we just let 4/x ->0 as x -> inf and we get the limit tending to x + 4?
the graph of $y = \sqrt{x^2+4x}$ is the upper branch of a hyperbola with a horizontal transverse axis ...
$y^2 = x^2+4x$
$y^2+4 = (x+2)^2$
$(x+2)^2 - y^2 = 4$
note the standard form for a hyperbola is $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$
$\dfrac{(x+2)^2}{2^2} - \dfrac{y^2}{2^2} = 1$ , a hyperbola centered at $(h,k) = (-2,0)$
Asymptotes have slope $m = \pm \dfrac{b}{a} = \pm \dfrac{2}{2} = \pm 1$
since the asymptotes pass through the hyperbola center ...
$y - 0 = -1(x + 2) \implies y = -x - 2$
and
$y-0 = 1(x+2) \implies y = x+2$