Why does an indeterminate cause a hole in a rational function and not a vertical asymptote? My intermediate Algebra textbook gave the following example and a graph of this function: 
$$f(x)=\frac{2x+1}{2x^2-x-1}$$

The factored form of this is:
$$\frac{2x+1}{(2x+1)(x-1)}$$ 
I know that a vertical asymptote is caused when a number is divided by $0$, since division by $0$ is undefined, and the vertical asymptote is located at the value of $x$ that causes a $0$ in the denominator. Therefore we set the factors of the denominator to $0$ and solve for $x$. 
$x-1=0$
$x=1$
And
$2x + 1 = 0$
$x=-\frac{1}{2}$
I can see why there's a vertical asymptote at $x = 1$, since $x =1$ will result in division by $0$, thus $x$ can approach $1$ but never be $1$ and as a result $y$ goes to either $\infty$ or $-\infty$ depending on which side $x$ approaches from.
However, taking $x=-\frac{1}{2}$ and plugging it back it will cause an indeterminate, but an indeterminate is still division by $0$, so shouldn't there be a vertical asymptote at $x = -\frac{1}{2}$ and shouldn't $y$ also be going to either $\infty$ or $-\infty$ as $x$ approaches $-\frac{1}{2}$? Why is it just a hole instead? 
 A: It is true that $\left| \frac{1}{2x^2 - x - 1} \right| \to +\infty$ as $x \to -\frac{1}{2}$.
However, it is also true that $2x+1 \to 0$ as $x \to -\frac{1}{2}$.
Your rule of thumb for identifying vertical asymptotes is basically the fact that, in the extended real numbers, you have $a \cdot (+\infty) = +\infty$ whenever $a$ is a positive number. (maybe the projective real numbers are more appropriate, since you're working with rational functions)
However, this particular example is a case of $0 \cdot (+\infty)$. This form is an undefined one (in the same way that $0/0$ is undefined), so this analysis can't tell you anything.

Everywhere near the value $x = -\frac{1}{2}$, you have
$$ \frac{2x+1}{2x+1} = 1 \qquad \qquad ({\small x \neq -\frac{1}{2}})$$
this expression is $1$ everywhere except at $x = -\frac{1}{2}$, so the graph of this expression is clearly just a hole.
So,
$$ \frac{2x+1}{(2x+1)(x-1)} 
= \left( \text{ 1 except there's a hole at $x=-\frac{1}{2}$} \right)
\cdot \left( \text{function well-behaved at $x=-\frac{1}{2}$ }\right) $$
and so its plot should be well-behaved except for a hole.
A: Consider this simpler example: $g(x)=\frac xx$. You have also a division by $0$ when $x=0$, but there's no vertical asymptote, since $g(x)=1$ for each $x\neq0$.
So, as in this example, an indeterminate doesn't always lead to a vertical asymptote. In your example, $f(x)=\frac1{x-1}$ when $x\neq-\frac12$ and therefore there is no vertical asymptote at $x=-\frac12$.
A: It's nothing more than the fact that $\lim_{x\to 1/2} f(x)$ is finite. For a rational function to have a vertical asymptote, the one-sided limits would need to be infinite (they needn't both be the same, one could be $+\infty$ and the other $-\infty$; compare $g(x)=1/x$ and $h(x)=1/x^2$ at $x=0$).
A: Simply,which value of x makes the numerator zero of any function is called zero point and which x value makes denominator zero means for what values of x the function becomes undefined is called poles.Here,we first have to check whether the numerator is zero or not.In this case,if the numerator would not be zero means at $x=\dfrac{1}{2}$ if 
$$f(x)=\dfrac{something}{0}~~~[something~~\ne~~0]$$
then we would surely call it a pole means there would be a vertical asymptote at $x=\dfrac{1}{2}$.But,it has also maked the numerator zero. So we have got $$f(x)=\dfrac{0}{0}$$ this is not possible...so it is undefined or there is no value for $f(x)$ there because this is not a valid expression.so,the function has a discontinuity there.
