Question

Explain how to find the equation for a vertical asymptote when given the equation.

Answer 1

rational functions

Answer 2

Put the denominator =0
Vertical asymptotes are at the values which make the denominator zero, that is x = ..

Answer 3

ok, what about discontinuities?

Answer 4

Hmm.. what do you mean?

Answer 5

explain how you find the location of a removable discontinuity(hole) in a graph when given the equation.

Answer 6

I...... also put denominator =0 ...

Answer 7

? both arent same though

Answer 8

Wait......
\[y= \frac{(x-1)(x+1)}{x-1}\]\(y= (x+1)\) where x ≠ 1

Answer 9

which is that?

Answer 10

Then, you draw a hole at x=1 for this example. => discontinuities

Answer 11

so let me get this straight,
(x+1) = asymptote
Whatever number that makes denominator=0 is the discontinuity?

Answer 12

x+1 is not the asymptote...

Answer 13

other way round?

Answer 14

Hmm.. nope...
For the function\[y= \frac{(x-1)(x+1)}{x-1}\]It can be reduced to y= (x+1). This is the one you graph.
The point is the you cancel the common factor x-1 in both numerator and denominator. You need to know that you can't cancel 0 as a common factor.
So, put x-1 =0, you get x=1
So, that's where the function is undefined. and the graph does not pass though the point at x=1. In other words, the graph is broken at x=1. => discontinuity

Answer 15

ok, so x=1 is discont (the number that makes the denominator 0)
and the vertical asymptote is?

Answer 16

I understand the discontinuity part

Answer 17

No vertical asymptote for that example.

Answer 18

well, how do i find the vertical asymptote in general?

Answer 19

A removable discontinuity occurs at the value of x that causes the function to 'equal' 0/0
A vertical asymptote occurs at the value of x that makes the denominator 0 but the numerator is still nonzero.

So basically, if the numerator and the denominator share a factor of x, then there is a 'removable discontinuity' at the value of x that causes that factor to be 0, i.e. you can 'remove' it like just 'a/a' (except you still have the restriction caused by that factor in the denominator after cancelling it).
If the denominator has a specific factor of x that causes only the denominator to be 0 (not in the numerator), then there is a vertical asymptote at that value.

Answer 20

ok, so when i cancel, what's left is the vertical asymptote?

Answer 21

I fail....

Answer 22

its okay, u still helped

Answer 23

I'm thinking...
When you can cancel the common factor, put the factor you cancel =0 to get the value for discontinuity.

When you see a fraction that cannot be simplified. Put the denominator =0 to get the vertical asymptote(s).

Answer 24

Yeah, that sounds good to me.

Answer 25

like this:
\[y=\frac{x^2 -x+1}{x-1}\] x=1 is the vertical asymptote

Answer 26

\[f(x)=\frac{\sum_k a_k x^k}{\prod_{j}(x-r_j)}\]
implies that there is a vertical asymptote at x=r_j for all j if and only if there aren't common factors in top and bottom, which there is a "hole" or "removable" discontinuity.

Answer 27

im only doing rational algebraic expressions

Answer 28

Tis what that is :P

Answer 29

\[\frac{a_k x^k + a_{k-1}x^{k-1}+...}{(x-r_j)(x-r_{j-1})...}\]

Answer 30

?????????????????

Answer 31

Alright i think i can summarize...
1. the number that makes the denominator=0 is the discont.
2. the vertical asymptote is what's left after canceling common factor
plz tell me if im wrong

Answer 32

Ehh... the common factor cancelled =0 fir discon.