Math Help - slant asymptote

1. slant asymptote

show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.

2. Originally Posted by johntuan
show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
A 'nasty' way (mathematicians turn away):

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

A much better way is to expand $\sqrt{x^2+4x}$ using the generalised binomial theorem and then take the limit.

3. haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function

4. Originally Posted by johntuan
haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function
Slant asymptotes are lines that the curve approaches as x --> +oo or x --> -oo. As x --> +oo or -oo, $(x+2)^2 - 4 \rightarrow (x+2)^2$ and the square root of this is $\pm (x+2)$. 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.