Is there a such thing as a rational function w/o a vertical asymptote?
364914
A rational function is the quotient of two polynomials.  A vertical asymptote only occurs when the denominator = 0.  So, any quotient of two polynomials such that the denominator is not = 0 would answer your question.
828415
You restrict the domain of the function as necessary in order that it pass both the vertical and horizontal line test.  Once you do that, f^-1 exists.  For example, if f(x) = sinx, you can restrict its domain to (-pi/2, pi/2). This function will then pass both h & v line test, and its inverse will then exist.\n\nanother example - f(x) = x^2.  A parabola opening upward whose vertex is the origin, and fails the h-line test.  If you restrict domain to (>=0),  f^-1 will then exist.
524475
It can be done using Taylor series (see first attached article)\n\nFunction f(x) can be approximated near some point x_0 by a sum of such series:\nf(x_0) + \n+ (x-x_0) * f'(x_0)/1! +\n+ (x-x_0)^2 * f''(x_0)/2! + \n+ (x-x_0)^3 * f'''(x_0)/3! + ...\n\nHowever, as you can see from section "List of Taylor series of some common functions" of first article, and section "Newton's generalized binomial theorem" of second article, binomial formula is correct not only for positive, but also for negative or even for complex p!\n\n\nThe only thing you need to do is generalize the concept of bnomial coefficients.\nUsually, they are counted as:\nC(n,m) = n!/(m!(n-m)!), but factorial are defined only for positive integers.\n\nUsing an alternative defiinition it becomes possible to operate with any real n:\nC(n,m) = n * (n-1) * (n-2) *...* (n-k+1) / k!.\nEasy to see, that for positive integer n those formulas are equivalent.
906736
AS you've been told, zero denominator corresponds to vertical asymptotes, so that gives x - 5 = 0, or x = 5 as one asymptote.\n\nThere are various ways of finding the horizontal one.  One intuitive method which is good in this case is ask yourself what sort of values do I get for f(x) if I put really huge values for x? \n e.g.  f(1000000) = (2000000 - 3)/(1000000 - 5) which is so close to \n2000000/1000000 that it doesn't matter, i.e. almost 2.  So your horizontal asymptote is y = 2.\n\nA formal way of doing this is to divide both top and bottom of the fraction by x, getting\n\n(2 - 3/x) / (1 - 5/x)\nAs x "approaches infinity"  (i.e. becomes boundlessly large), both 3/x and 5/x approach zero, and so f(x) approaches 2/1, so the asymptote is y = 2.\n\nYet another method is to divide 2x -3 by x - 5, getting 2 with remainder 7, and so\n\nf(x) = 2  +  7/(x - 5), and the larger x becomes, the smaller 7/(x-5) becomes, and so f(x) approaches 2.
280197
Asymptotes are linear areas of the graph where no values will satisfy the equation - in other words, the value of the function approaches either 0 or +_infinity.\n\nObviously, x can never be 0, so x = 0 would be a vertical asymptote. But for any non-zero value for x, y can be just about anything. However, from the function, you can see that the numerator is a constant, so y will never reach 0, no matter how large x gets. So y = 0 is a horizontal asymptote.
939642
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
813038
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
740420
Wait a minute everybody!\nthose vertical lines,( lyl ) could mean absolute value!!\nHelloooo?\nso the answerSSS could be 9 and -9\nsheesh!
1154261
The third answer above is close, but not quite right.  A function can have two different horizontal asymptotes -- one in the limit of +infinity, and the other in the limit of -infinity.  You need to check the values of both limits separately.  \n\nFor instance, the horizontal asymptotes of arctan(x) are +pi/2 (as x -> +infinity) and -pi/2 (as x -> -infinity).\n\nFor rational functions (functions that can be expressed as the ratio of two polynomials) there is an easy way to test for and find the horizontal asymptotes:\n\nLet  f(x) = p(x)/q(x) be a rational function. Then: \n\n1.    If (the degree of q(x))  > (the degree of p(x)) then there is one horizontal asymptote at 0.\n\n\n2.    If (the degree of q(x)) =  (the degree of p(x)) then there is one horizontal asymptote at the (leading coefficient of p(x))/(leading coefficient of q(x))\n\n\n3.    If (the degree of p(x)) < (the degree of q(x)) then there is no horizontal aymptote.
957617
2d point of no return
860906
it is called akermen function. search web about this
570055
Vetrtical asymptotes occur when the denominator is zero.  5x^2 - 25x = 5x(x-5)=0 if x = 0 or 5, so vertical asymptotes are x=0 and x=5.\n\nHorizontal asymptotes occur when the top and bottom have the same degree (highest exponent).  Just take the coefficients of the terms with highest exponent.  So horizontal asymptote is y = 11/5\n\n(They also occur if the top degree is less than the bottom, at y=0).
411240
you get vertical asymptote when denominator of a fraction hits zero. so \nx=0 for 3/2x\nx=-1 and x=3 for (x/(x+1)(x-3))\nand so on
163263
The only functions that will qualify for an oblique asymptote are rational functions, in other words fractions. In addition, one more thing has to happen--the degree of the numerator has to be exactly one more than the degree of the denominator. Basically, the highest power in the top of the fraction has to be 1 more than the highest power in the bottom.\n\nTo find the asymptote, you need to use long or synthetic division to divide the top by the bottom. Once you do, discard the remainder and set the quotient equal to y; that will be the linear equation of the asymptote.\n\nI know that may sound confusing, so I worked out an example for you. You need visualize this to understand it, so click the links below in "sources" to see the answer graphic hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nThe first link is the graph with its asymptote and the second shows the steps you follow to get that asymptote.
826336
y=6/ln(x)-9\nput x=e^9,y=infinity\nso x=e^9 is vertical asymptote\nlnx=(6/y)+9\nx=e^[(6/y)+9]\nx is not equal to infinity for any value of y...\nso no horizontal asymptote\n\n\nhope that helps
978117
The graph of f(x) is a line, which means it can't have an asymptote.\n\ng(x) = log5(2x) should have a vertical asymptote x = 0.
424740
Unfortunately there is not. There is however an algorithm that will allow you to work out square roots by hand.