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:

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


thank you in advance

let's start with something simple to see what you know ...



where are the vertical asymptotes?

what is the horizontal asymptote?

where is the point discontinuity (hole) ?

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

Quote:

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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

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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



horizontal asymptote at y = ?

two vertical asymptotes ... where?

one "hole" ... where?

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?

horizontal asymptote is y = 0 ... the degree of the numerator < degree of the denominator

vertical asymptotes correct

hole at x = 0

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?

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 has a "hole" at . the graph of the function will look like the line except for the "hole" at ...

note the graph made on my TI calculator ... see the "hole" ?

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?

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).

so if there are holes, are the common factors canceled first? then there would be no Vertical asymptotes?

the function has both a hole at one x-value and a vertical asymptote at another x-value ... where?

hole at 1/2. but would 3 and -1 still be verticle asymptotes and would -1 still be a root?

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

ok i see now, thank you very much!