# Why the vertical asymptote of y=f(x) is the zeros of $1/f(x)$?

Why the vertical asymptote of $y=f(x)$ is the zeros of $1/f(x)$? For example, if $f(x)$ has a vertical asymptote $x=1$, then $(1,0)$ must be a x-intercept of $1/f(x)$?

I saw this in IB textbook and I want to know why.

the vertical asymptotes of $y = f(x)$ become zeros of $y = \frac{1}{f(x)}$

• Your statement is a little vague and possibly false. Can you give a specific $f(x)$ that you're thinking about? – Matthew Leingang Nov 1 '17 at 15:11
• For $f(x) = \frac{z^2+2z+1}{z+1}$ this is not the case, as there are no vertical asymptotes. – Martin Nov 1 '17 at 15:34
• Often $f(x)$ is undefined at $a$ if the line $x=a$ is a vertical asymptote, and then $1/f(x)$ is also undefined at $a$, so strictly speaking the claim doesn't make sense. (But one might get a removable singularity, of course.) And if $f(a)$ is defined, then definitely $1/f(a)$ can't be zero... – Hans Lundmark Nov 1 '17 at 15:39
• By the way, what is “IB textbook”? – Hans Lundmark Nov 1 '17 at 15:40
• @HansLundmark IB = International Baccalaureate. It's an international secondary school curriculum. – Matthew Leingang Nov 1 '17 at 16:06

It sounds like you're saying this:

If the graph of $f(x)$ has a vertical asymptote at $x=a$, then $1/f(a)=0$.

Example: Let $f(x) = \frac{x+1}{x-1}$. This has a vertical asymptote at $x=1$. The reciprocal expression is $\frac{x-1}{x+1}$, and the associated function has a zero at $x=1$.

But this is being very sloppy about the relation between a function and its reciprocal. The reciprocal of $f(x)=\frac{x+1}{x-1}$ is $\frac{1}{f(x)} = \frac{x-1}{x+1}$, but the domain of $f$ excludes $1$. So actually $$\frac{1}{f(x)} = \begin{cases} \frac{x-1}{x+1} & x \neq 1 \\ \text{undefined} & x=1 \end{cases}$$ So there is no value $a$ for which $\frac{1}{f(a)} = 0$.

And come to think of it, the equation $\frac{1}{f(a)} = 0$ is already problematic, because if it were satisfied, then $$\frac{1}{f(a)} = 0 \implies 1 = 0 \cdot f(a) = 0.$$ And this is a contradiction.

We can extend $f$ to give it a value at its asymptote, in such a way that the reciprocal is defined. But that reciprocal value still won't be zero. For instance, suppose we said $$f(x) = \begin{cases} \frac{x+1}{x-1} & x \neq 1 \\ 2 & x=1 \end{cases}$$ Then, $$\frac{1}{f(x)} = \begin{cases} \frac{x-1}{x+1} & x \neq 1 \\ \frac{1}{2} & x=1 \end{cases}$$ So $f$ has an asymptote at $x=1$, and $\frac{1}{f}$ is not zero at $x=1$.

I am wondering why this was written in your textbook when it's so wrong. It could be that only a certain class of functions are being considered. For instance, I think it's true for rational functions $f(x) = \frac{p(x)}{q(x)}$, where $p$ and $q$ are polynomials. This $f$ has vertical asymptotes when the denominator tends to zero and the numerator to something other than zero. That is, if $q(a) = 0$ and $p(a) \neq 0$, then $f$ has a vertical asymptote at $a$. Then the reciprocal expression $f(x) = q(x)/p(x)$ does have a zero at $a$. But this isn't quite the same thing as the function $\frac{1}{f(x)}$.