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? 

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.
