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:
https://www.desmos.com/calculator/px9t5ap0lo
(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?
Best Answer
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.
Actually.
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 $$\lim_{x\rightarrow-3^-}\frac{(x+1)(x-1)}{(x-1)(x-2)(x+3)}=-\infty.$$