Question

Can someone give me a quick explanation of vertical asymptotes, horizontal asymptotes, removable discontinuities, holes, and non removable discontinuities.

Answer 1

Vertical asymptotes show up in rational functions.
They are determined based on the information we have in the denominator:\[\large\rm y=\frac{x^2+4x+3}{(x-2)}\]In this example, notice that the denominator is zero exactly when x=2.
So we have a vertical asymptote at x=2.

Answer 2

We have a removable discontinuity when that factor which would normally give us a vertical asymptote can be removed from the equation:\[\large\rm y=\frac{x^2+4x+3}{(x+1)}\]So when you look at this new example, you might think that we have a vertical asymptote at x=-1. But the numerator can be factored,\[\large\rm y=\frac{(x+1)(x+3)}{(x+1)}\]And from this point we can see that the (x+1) factor can be `removed` or `cancelled out`.

Answer 3

So by removing the factor, we remove all of the asymptotic behavior that's happening around x=-1. But we still have the restriction that x cannot equal -1 because (x+1) was a part of the original equation.

Answer 4

ok, so should I factor first?

Answer 5

Yes, when dealing with rational expressions,
factor before deciding how to label this and that :)

Answer 6

and is a vertical asymptote a hole?

Answer 7

oh, wait a hole is a removable discontinuity right?

Answer 8

Yes, a removable discontinuity leaves a hole behind.

Answer 9

another question, how do you find out if it is a vertical or horizontal asymptote

Answer 10

They're very different things.

Denominator tells us about vertical asymptotes.
So we have specific x values that tell us about vertical asymptotes.

Horizontal asymptotes occur wayyyy wayyy off in the distance.
On the far left and right of the graph.

When x is really really really big, what is the function doing?
Is it growing out of control?
Or
Is it leveling off at some height? (This is a horizontal asymptote).

Answer 11

leveling off at some height, horizontal asymptote

Answer 12

Here is a quick example:\[\large\rm \frac{1}{(x-1)(x+2)}\]We get a zero in the denominator whenever (x-1)=0 or (x+2)=0.
So we have vertical asymptotes at x=1 and x=-2.

Alternatively, looking for horizontal asymptotes,
we can plug a really big x-value into this function and see,\[\large\rm \frac{1}{(2bajillion-1)(2bajillion+2)}=\frac{1}{a~lot}\]

Answer 13

1/(a lot) is approximately zero, it's a really really tiny fraction.

Answer 14

So we can see that this particular example has a horizontal asymptote at y=0.

Answer 15

It's leveling off at y=0 when x gets really really big.

Answer 16

going back, removable discontinuities are when you can factor out the denominator and numerator, while a nonremovable is when....

Answer 17

I'm not sure.
I've never heard the term `non-removable discontinuity` before.
I guess maybe it's describing all other types of discontinuities.
Everything that is NOT a removable discontinuity.

This would include:
~Asymptotic discontinuities
~Jump discontinuities

Or maybe it just describes asymptotes, the first one.
Sorry I've never heard that term before D:

Answer 18

ok, well thank you for all the other help, I appreciate it :)

Answer 19

np