Why does $\frac{x^2-4x+3}{x^2-1}$ have only one asymptote?

Question

I want to find the asymptotes of the function

$f(x)=\frac{x^2-4x+3}{x^2-1}$.

I feel like I am missing something here... But I understand that the denominator should factor out into $(x+1)(x-1)$ and therefore return two asymptotes. Why doesn't it?

Answer 1

The numerator factors and cancels with one of the factors in the denominator:

$f(x)=\frac{x^2-4x+3}{x^2-1}=\frac{(x-3)(x-1)}{(x+1)(x-1)}=\frac{x-3}{x+1}$.

Hence, we have a vertical asymptote at $x=-1$ and a hole in the graph at $x=1$. Then we have a horizontal asymptote at $y=1$.

Answer 2

The numerator factors out into $(x-1)(x-3)$ so the function can be simplified into $$\frac{x-3}{x+1}.$$

Answer 3

While I would like to echo what has already been said, namely that the function can be written as $$f(x) = \frac{(x-3)(x-1)}{(x+1)(x-1)},$$ it is not true that $$f(x) = \frac{x-3}{x+1}$$ as has been mentioned because in the first equation $x=1$ is not in the domain and in the second equation $x=1$ is in the domain. Technically the cancellation can happen except for at $x = 1$, where it should be noted that the function remains undefined. Therefore $$f(x) = \frac{x-3}{x+1}\quad (x\neq 1)$$ is the appropriate simplification. A subtle distinction, yes, but an important one. What that tells us is that $x = -1$ is a vertical asymptote and at $x = 1$ there will be a hole in the graph. In fact, by plugging in $x = 1$ to $\frac{x-3}{x+1}$ we can see where the point would be, so the graph has a hole at the point .$(1,-1)$