1. ## Discontinuity question

The graph of y=(x^2-9)/(3x-9) has:
(sorry for not typing out the answer not related to my question)
a) a vertical asymptote at x=3
b)....
c) a removable discontinuity at x=3
d)...
e)...

I know the answer is c but I got confused with answer a.
Doesn't the graph of y=(x^2-9)/(3x-9) has a vertical asymptote at x=3 if numbers are plugged into the equation? How do we know if (x-3) should be factored out or not?
Thanks

2. ## Re: Discontinuity question

You need to recognise that for all $\displaystyle x \neq 3$ that $\displaystyle \frac{x^2 - 9}{3x - 9} \equiv \frac{(x - 3)(x + 3)}{3(x - 3)} \equiv \frac{x + 3}{3}$.

Since the function is identical to a linear function, which does not have asymptotes, that means that x = 3 is not an asymptote, it's just a "hole", or a removable discontinuity.

3. ## Re: Discontinuity question

Originally Posted by Prove It
You need to recognise that for all $\displaystyle x \neq 3$ that $\displaystyle \frac{x^2 - 9}{3x - 9} \equiv \frac{(x - 3)(x + 3)}{3(x - 3)} \equiv \frac{x + 3}{3}$.

Since the function is identical to a linear function, which does not have asymptotes, that means that x = 3 is not an asymptote, it's just a "hole", or a removable discontinuity.
Thanks a lot!