How do asymptotes work.. in layman's terms...?
108513
Asymptotes are vertical lines associated with a graph but not part of a graph.  They are divisions in a graph that I call "territory markers" or "fences".  They are essentially the x values that make any function undefined.  These x values, when graphed, are vertical lines that a graph will  not cross but rather be forced to "walk along".\n\nExample:  f(x)=1/((x)*(x-2))\n\nThe function is obviously undefined at 2 specific x values (set the denominator equal to zero and solve for those x vales to find where the function is undefined), x=0 and x=2.  The graph of the line x=0 and/or x=2 is the graph of a vertical line, or asymptote, that this function cannot cross directly since there is no corresponding y value at that point (the function is undefined at these points, remember?).  The function is forced to skirt the fence line either up to infinity or down to negative infinity.  Graph it and you'll see.
742119
asymptote at x=5 and y=2
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.
1144759
I don't know the answer to your question, but here is a good website that might help:\n\nhttp://www.purplemath.com/modules/asymtote3.htm
963822
The vertical asymptotes occur where you are evaluating y = tan(pi/2 + k pi):\n\n2x - pi/4 = pi/2 + k pi\n2x = 3pi / 4 + k pi\nx = 3 pi / 8 + k pi / 2\n\nWithin the domain 0 <= x <= 2pi, this gives:\n\nx = 3 pi / 8, 7 pi / 8, 11 pi / 8, or 15 pi / 8
712145
f(x) = ba^x + c\nSince it passes through the point P(1,4)\nf(1) = ba + c = 4\nSince the y-intercept is 2 (ie it passes through (0, 2)\nf(0) = b + a = 2\nThe statement "horizontal asymptote y=2" renders a solution impossible. I think you meant horizontal asymptote y=1\nThen lim x→-∞  f(x) = c → c = 1\nSo b + 1 = 2 → b = 1\nba + c = 4 → a = 3\n\nThus f(x) = 3*x + 1
921732
If you look at a picture (aka a graph) of the function, it will all become clear...\n\nhttp://upload.wikimedia.org/wikipedia/en/3/3e/1-over-x.png
939596
Since we don't have any translations here, the general form is:\nx^2/a^2 - y^2/b^2 = 1.\na is the position of the vertices; ie a = 3.\nThe formula for the asymptotes is y = + or - (b/a)x. So b/3 = 1/2, so b = 1.5.\nSo x^2/3^2 - y^2/1.5^2 = 1.
909100
To have vertical asymptotes at x = 6 and x = -1, the denominator should have factors of (x - 6) and (x + 1), making it x^2 - 5x - 6.  To have a horizontal asymptote of -9, the numerator should be of the same order as the denominator, but with a coefficient for the greatest exponent of x that is -9 times that of the denominator.  The denominator is 2nd order with a coefficient of 1 for x^2, so the numerator can be -9x^2.\n\n-9x^2 / (x^2 - 5x - 6)
164065
The trick with "proofs" like these are generally an obscured division by zero. One of the lines seems to work, though when you divide to cancel out some factors, the quantity you're dividing by could/does equal zero. It's a trick and doesn't say anything substantial about mathematics.
1112728
to draw lines u hav to select 2 points at a time (as no 3 r collinear)\n\nthus no of selections of such points is nC2= n!/ [ 2! (n-2)!]
1147966
Yes!!!!!!!!! Email me, and I'll try to shed more light on this. Parabolas are where my genius comes out :)
1075799
The vertical asymptote is the equation x=6.\n\nThe horizontal asymptote is R'(x). Therefore, the horizontal asymptote is the equation y=3.\n\nThis function does not have oblique asymptotes.
946291
Vertical asymptotes will occur where the function is undefined. Usually this happens when you have a division by 0. In this case, when x = 0 you can't evaluate the function, and as it turns out, there is a vertical asymptote there.\n\nHorizontal asymptotes are found by taking the limit of the function as x goes to either positive or negative infinity. In this case, when x gets really big, the function gets really close to 1. As it turns out that happens for both negative and positive infinity.\n\nSo the answer is vertical asymptote at x = 0 and horizontal asymptote at y = 1.
966227
No oblique assymptotes, since the highest degree in the numerator is the same as in the denominator.\n\nlet's factor both equations:\n\n2x^2-x-10 = (2x-5)(x+2)\n-x^2+2x+24 = (6-x)(x+4)\n\nNow that we have that done, it will be easier.  There are no zeroes in the numerator that the denominator shares, which means that all zeroes of the denominator will be vertical assymptotes.\n\nx=6\nx=-4\n\nFor horizontal assymptotes, multiple the top and bottom by 1/x^2 (highest degree of x).  We get\n\n(2-1/x-10/x^2)/(-1+2/x+24/x^2) As x approaches infinity, this approaches 2/-1 = -2\n\nThus our horizontal assymptote is y=-2
993077
not that I know...you might have to wright your own algorithm!\n\nouch...have fun!\n\nbut check links!!
777035
a) \nThe reciprocal of a is y =1/(5x-10)\n\nThe horizontal aysmptotes is the value which the graph tends towards but never reach. This is done by placing a extremely large number.\n\nYou don't actually place a large number but think if x is really large y becomes like y=1/5x as -10 will not make a difference if x is really large\n\n5*"this really large number" would be even bigger so y becomes y=1/"a really big number"\n\nWhen you divided by a bigger number it always gets closer and closer to 0 but never reaches it\n\nSo the asymptote is y=0\n\nb)\n\nReciprocal is y=1/(x^2-x-20)\n\nYou do the same thing\n\nAs x^2 is the dominate power, when x ->infinity y = 1/(x^2)\n\nAgain the denominator is larger than the numerator and so the graph will tend towards 0\n\nAgain the horizontal asymptote is y=0