Question

Okay so we're studying Vertical Asymptotes...
If f(x)=(x+2)/(x^-2x-3)
The asymptotes are 3 and -1, right?
Then would the domain be "X is all real numbers except for -1,3"?

But on the GDC it only shows 3 as an asymptote...?

Answer 1

-1 is an asymptote

Answer 2

So the domain would be "X is all real numbers except for -1 and 3"?

Answer 3

if there are like factors on top and bottom; they cancel out and create a "jump", or a hole.

after canceling out like terms; if there is anything left on the bottom, the value that zeros it out will be an asymptote\[\frac{ab}{bm}\]since the bs cancel, they create a jump discontinuity. After canceling the bs, we are left with an "m" in the bottom which would create an asymptote

Answer 4

the domain would still exclude b and m regardless

Answer 5

Thank you!

Answer 6

but there is nothing to cancel in this problem
\[
\frac{x+2}{x^2-2 x-3}=\frac{x+2}{(x-3) (x+1)}
\]

Answer 7

so x=3 and x=-1 are both vertical asymptotes.