
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/cgiperl/sy...28x%20%203%29
thank you in advance

Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
let's start with something simple to see what you know ...
$\displaystyle y = \frac{(x1)(x+2)(x+3)}{2(x+1)(x+2)(x3)}$
where are the vertical asymptotes?
what is the horizontal asymptote?
where is the point discontinuity (hole) ?

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

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
$\displaystyle g(x) = \frac{6x^2}{x^4  4x^2} = \frac{6x^2}{x^2(x^2  4)} = \frac{6x^2}{x^2(x2)(x+2)}$
horizontal asymptote at y = ?
two vertical asymptotes ... where?
one "hole" ... where?

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?

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

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?

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Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
A "hole" is really a point discontinuity ... the discontinuity occurs at the xvalue that makes the common factor in the numerator and denominator equal to zero.
for example, the function $\displaystyle y = \frac{x^21}{x1} = \frac{(x+1)(x1)}{x1}$ has a "hole" at $\displaystyle x = 1$. the graph of the function will look like the line $\displaystyle y = x+1$ except for the "hole" at $\displaystyle x = 1$ ...
note the graph made on my TI calculator ... see the "hole" ?

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?

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).

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?

Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
the function $\displaystyle y = \frac{(x+1)(x2)}{(x3)(x+1)}$ has both a hole at one xvalue and a vertical asymptote at another xvalue ... where?

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?

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 yvalue of the hole is 3/4
there is one vertical asymptote at x = 3

Re: Finding Verticle asymptotes, holes and, and horizontal asymptotes
ok i see now, thank you very much!