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?
Best Answer
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.