Curved sketching strange things happen

[rock.freak667]

I was told to sketch this graph:

$$y=\frac{x-1}{x^2-100}$$

when x=0,y=.01


$$y=\frac{\frac{1}{x}-\frac{1}{x^2}}{1-\frac{100}{x^2}}$$

as $x \rightarrow \infty;y \rightarrow 0 \Rightarrow$ y=0 (i.e. x-axis is a horizontal asymptote to the curve)

BUT

when y=0; x-1=0 so that x=1. So a point on the curve is (1,0)

How can the x-axis be a horizontal asymptote to the curve yet the curve passes through (1,0)?

 

[uman]

What do you mean, how can it? It just is.

The x-axis is an asymptote for $$sin(x)/x$$, and this is equal to zero an infinite number of times.

 

[Vid]

There's a vertical asymptote at x=10. For x > 10, the x-axis is a horizontal asymptote.

 

[rock.freak667]

There's a vertical asymptote at x=10. For x > 10, the x-axis is a horizontal asymptote.

(fixed)


OH...that clears it right up for me! Thank you. I was under the impression that the x-axis would be a horizontal asymptote for the entire graph no matter what.

 

[Tedjn]

It is a horizontal asymptote for the entire graph. The notion of horizontal asymptote just means what the function f(x) approaches something as x approaches positive or negative infinity. It has nothing to do with f(1), and as you have found, it is quite possible to cross over an asymptote. See uman's post for a good example of a such a function.