Answer to Question #97894 in Calculus for ALexa

Question

In a very interesting case, it is sometimes possible for a graph to have a horizontal asymptote, but to have a location where the graph crosses that "border." For example: look at this
(Links to an external site.)
Desmos graph for the function
f ( x ) = 2 x 3 x 2 + 1
. Note that it has a horizontal asymptote at
y = 0
, but also notice that the graph does cross that line, right at the origin.
Do you think that it is possible for a graph to cross a vertical asymptote? Why or why not?

Answer 1

The function $ƒ(x)$ may or may not be defined at a, and its precise value at the point $x=a$ does not affect the asymptote. For example, for the function:

$f(x)=1/x; \forall x>0$

$=5; \forall x<=0$

Here the function has a limit of $+∞$ as $x → 0^+$.

$\therefore ƒ(x)$ has the vertical asymptote $x = 0$ even though $ƒ(0) = 5$ .

The graph of this function does intersect the vertical asymptote once, at $(0,5)$ . But at the same time, it is worth noting that the function has a discontinuity at $x=0$ .

It is impossible for the graph of a function to intersect a vertical asymptote (or a vertical line in general) in more than one point. Moreover, if a function is continuous at each point where it is defined, it is impossible that its graph does intersect any vertical asymptote.

Conclusively, it can be said that for a graph to cross a vertical asymptote, it's function must be discontinuous.

Graph of a continuous function can never cross a vertical asymptote due to the aforementioned reasons.