Here's the question you clicked on:
mamma17
Describe the vertical asymptote(s) and hole(s) for the graph of y=(x-1)(x+3)/(x+3)(x-5) [1:06:17 PM] Akora: asymptote: x = –1 and hole: x = 5 asymptote: x = 5 and hole: x = –3 asymptote: x = –5 and hole: x = 3 asymptotes: x = 5 and x = –3
why did you close the last question? It looks like you're pasting the questions from a test.
no it aint from a testitis jux i did not neeit any more ani figured it out and copy them from skype cause i am asng some oer fiends as well
oh? so what was the answer to the last question?
OK that's right. If you're really not cheating, I'm sorry I doubted you.
iaint it is a quick checkand idid the lesson jux d not undrsand it
\[ y=\frac{(x-1)(x+3)}{(x+3)(x-5)}\]
ok then. This question is exactly like the previous questions. Do the exact same thing.
you can cancel the \(x+3\) factor, which means it has a "hole" at \(x=-3\)
however you cannot cancel the factor of \(x-5\) so it has a vertical asymptote at \(x=5\)