How can there be an x and y intercept of (0,0) if there is an asymptote at y=0?

Question

anonymous · Answer

For example, in this rational function$$\frac{ x }{ x^2-1 }$$
there is a horizontal asymptote at y=0, but there is still the intercept of (0,0). How can this be?

geerky42 · Answer

I suggest you to try and graph it

phi · Answer

It's not a straight line.  It curves

anonymous · Answer

As you can see from the graph, it passes through the origin, yet there is an asymptote at y=0. I don't understand how this can happen. http://www.wolframalpha.com/input/?i=%28x%29%2F%28x%5E2-1%29

phi · Answer

you have
$$ y = \frac{x}{x^2 - 1} $$

when x is zero you get
$$ y= \frac{0}{-1} = 0 $$

when x is a very large number, x^2 - 1 is (almost) x^2
(compare   100^2  to 100^2 -1 , they are close in value)

so for large positive x we can say x/(x^2-1) is about x/x^2  = 1/x
and as you know for large x, 1/x approaches 0
i.e.  y ->0  as x ->infinity

so this function does go through (0,0) and it asymptotes to y=0

that is what non-linear expressions do:  complicated things

anonymous · Answer

okay, thank you for that explanation!