# Finding Verticle asymptotes, holes and, and horizontal asymptotes

• Dec 5th 2012, 02:07 PM
Recyclop
Finding Verticle asymptotes, holes and, and horizontal asymptotes
im having a really difficult time finding out how to find the verticle asymptote and holes and horizontal asymptote. its really frustrating (Headbang)

for exmaple here are some questions:

http://www.webassign.net/latexImages...619c1726cf.gif

http://www.webassign.net/latexImages...cdd4b9eb44.gif

http://www.webassign.net/cgi-perl/sy...28x%20-%203%29

• Dec 5th 2012, 03:09 PM
skeeter
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes

$y = \frac{(x-1)(x+2)(x+3)}{2(x+1)(x+2)(x-3)}$

where are the vertical asymptotes?

what is the horizontal asymptote?

where is the point discontinuity (hole) ?
• Dec 5th 2012, 06:15 PM
Recyclop
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
out of my little understanding of it (my instructor sucks)
i can say that
horizontal asymptote at 1/2
vertical asymptotes at -1,-2, and 3
holes at........i dont even know. LOL
• Dec 5th 2012, 06:40 PM
skeeter
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
Quote:

Originally Posted by Recyclop
out of my little understanding of it (my instructor sucks)
i can say that
horizontal asymptote at 1/2
vertical asymptotes at -1,-2, and 3
holes at........i dont even know. LOL

horizontal asymptote at y = 1/2 ... it's the equation of a line

vertical asymptotes at x = -1 and x = 3 , "hole" at x = -2 ... note the common factor in the numerator and denominator.

OK ... using one of your problems

$g(x) = \frac{6x^2}{x^4 - 4x^2} = \frac{6x^2}{x^2(x^2 - 4)} = \frac{6x^2}{x^2(x-2)(x+2)}$

horizontal asymptote at y = ?

two vertical asymptotes ... where?

one "hole" ... where?
• Dec 5th 2012, 08:34 PM
Recyclop
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
so horizontal asymptote @ y=6
verticle asymptotes at x=+2 and x=-2
the hole would be at.....X^2 would be the only common factor i see. but that would be sqrt0.... so there is no hole?
• Dec 6th 2012, 04:09 AM
skeeter
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
horizontal asymptote is y = 0 ... the degree of the numerator < degree of the denominator

vertical asymptotes correct

hole at x = 0
• Dec 6th 2012, 03:49 PM
Recyclop
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
so after class today i got a good understanding of asymptotes and what not.
only thing that still doesnt make sense is holes. i understand that a hole occurs when you have common factors but what do i do to find the point and what does a hole mean?
• Dec 6th 2012, 04:36 PM
skeeter
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
A "hole" is really a point discontinuity ... the discontinuity occurs at the x-value that makes the common factor in the numerator and denominator equal to zero.

for example, the function $y = \frac{x^2-1}{x-1} = \frac{(x+1)(x-1)}{x-1}$ has a "hole" at $x = 1$. the graph of the function will look like the line $y = x+1$ except for the "hole" at $x = 1$ ...

note the graph made on my TI calculator ... see the "hole" ?
• Dec 6th 2012, 07:42 PM
Recyclop
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
ya i see, so how do we determine the y value? is it just whatever number happens to be there if we plug in 1 for the complete function?
• Dec 7th 2012, 04:04 AM
skeeter
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
sub in x = 1 after dividing out the common factor ... in this case, y = x+1 = 1 + 1 = 2 ... the "hole" is at (1,2).
• Dec 7th 2012, 08:46 AM
Recyclop
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
so if there are holes, are the common factors canceled first? then there would be no Vertical asymptotes?
• Dec 7th 2012, 11:22 AM
skeeter
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
the function $y = \frac{(x+1)(x-2)}{(x-3)(x+1)}$ has both a hole at one x-value and a vertical asymptote at another x-value ... where?
• Dec 7th 2012, 12:03 PM
Recyclop
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
hole at 1/2. but would 3 and -1 still be verticle asymptotes and would -1 still be a root?
• Dec 7th 2012, 01:43 PM
skeeter
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
sorry, but no ...

there is a hole at x = -1 (the value of x that makes the factor common to the numerator and denominator 0), the y-value of the hole is 3/4

there is one vertical asymptote at x = 3
• Dec 7th 2012, 01:56 PM
Recyclop
Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
ok i see now, thank you very much!