I have this function: $\frac{x^2-1}{x^3-7x+6}$ and I need to find its x-intercepts and vertical asymptotes.

I factored it as: $\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}$.

Is the graph produced by desmos correct:


(if not, which online graph tool produces correct graphs?)

I think it is "almost" correct, because it just needs to additionally show that at x=1 function is not defined (as a point, not as an asymptote).

Vertical asymptotes are at x=-3 and x=2, right?

  • $\begingroup$ @postmortes are you sure the function is defined at $x=1$? Certainly you can extend the function to be defined at that point using the equal LH/RH limits, but wouldn't the natural domain of the function exclude $1,2$ and $-3$? $\endgroup$ – BaroqueFreak Aug 19 at 14:18
  • $\begingroup$ @postmores, that is NOT true. The limit of a function at a given x does not have any necessary relation to the value of the function there. For example the function defined as "f(x)= 3x+ 1 for x not equal to 1, f(1) not defined" has limit 4 at x= 1 but is not defined there. Of course, it is not continuous there. There was no condition here that the function be continuous. $\endgroup$ – user247327 Aug 19 at 14:57

You are right.

I think, it's better to write the following.

$$\lim_{x\rightarrow2}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=\infty,$$ which says $x=2$ is an asymptote. $$\lim_{x\rightarrow-3}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=\infty,$$ which says $x=-3$ is an asymptote. $$\lim_{x\rightarrow1}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=-\frac{1}{2},$$ which does not give asymptote.


We don't need to write here: $$\lim_{x\rightarrow2^+}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=+\infty,$$ $$\lim_{x\rightarrow2^-}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=-\infty,$$ $$\lim_{x\rightarrow-3^+}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=+\infty$$ and $$\lim_{x\rightarrow-3^-}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=-\infty.$$


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