what is the vertical asymptote of  X^3/ X^2 + 2X - 15?
819082
Let f(x) = x ³/ (x² + 2x - 15) \n\n= x - 2 + (19x -30)/(x² + 2x - 15) (by division)\n\n= x - 2 + (19x -30)/[(x - 3)(x + 5)]\n\nSo f(x) is undefined when x = -5, 3\n\nAlso lim x→±∞ f(x) = x - 2\n\nThus f(x) is asymptotic to g(x) = x - 2 as x →±∞ and asymptotic to x = -5 and x = 3
908882
There are three types of asymptotes: horizontal, vertical and slant.  \nVertical asymptotes occur when the denominator is zero (example: y = 3/(x-1) has a v.a. at x = 1).\n\nHorizontal asymptotes occur when the degree of the top is less than the degree of the bottom, in which case the h.a. is y = 0.\n\nOR, when the degree of the top is equal to the degree of the bottom, in which case the h.a. is\n y = (leading coefficient of top)/(leading coefficient of bottom)\n\n(example: y = (3x+6)/(x-5) has a h.a. at y = 3/1 or y = 3)\n\nIt has a slant asymptote if the degree of the top is exactly one more than the degree of the bottom.  To find its equation, divide the top by the bottom and take the quotient (disregard the remainder).  Note: if there is no remainder, you should have reduced the rational expression first.\n\n(example: y = (x^2 + 3x + 4)/(x - 5) has a s.a. at y = x + 8\n\nI would be happy to explain more if you want, just send an IM
1005028
if \na=(v-u)/t\nthen\na*t=v-u\nthen\nt=(v-u)/a\nor\nt=(15-0)/0.75 \n\nSo as the previous poster says the answer really is 20
787424
To find the asymptote to the left.  find the limit of f(x) as x approaches -∞.\n\nTo find the asymptote to the right.  find the limit of f(x) as x approaches ∞.\n\nVertical asymptotes usually occur where a value of x gives a division by zero.  Find when the denominator equals zero then take the limit as x approches that value to make sure the value of the function approaches negative or positive infinity.\n\n I hope that helps.
216656
(x^4 - 25) = (x^2 - 5)(x^2 + 5) = (x - 5)(x + 5)(x^2 + 5)\n\n(x^4 - 25)/(x - 5) = [(x - 5)(x + 5)(x^2 + 5)]/(x - 5)\n(x^4 - 25)/(x - 5) = [(x + 5)(x^2 + 5)]\n(x^4 - 25)/(x - 5) = x^3 + 5x^2 + 5x + 25\n\nBut remember that division by 0 is a no-no, so looking at the original expression, a graph of this will have a vertical asymptote at x = 5.
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
768619
the horizontal asymptote is the limit as x->∞, the vertical as x->0.  Since this equation can be written -9 + 1/x^2, when x gets large, 1/x^2 gets small with respect to -9, and can be ignored, leaving  -9 as the symptote.   As x->0, the 1/x^2 dominates, so the vertical asymptote is 1/x^2.  There is a discontinuity at x=0 where the function is undefined.
465898
sqrt(20/3)\n(sqrt(20))/(sqrt(3))\nMultiply top and bottom by sqrt(3)\n(sqrt(60))/3\n(sqrt(4 * 15))/3\n(2sqrt(15))/3
742119
asymptote at x=5 and y=2
921915
NOTE: i know this looks long, but i put in a lot of explanation to help on this problem and any others like it. i think (and hope) that youll find very useful! :)\n\n\nyoure right about the vertical asymptote(VA), you just set the denominator equal to zero:\n\nx(x-1) = 0\n\nyou solve it using the zero product theorem.  that says that when the product of two things equals zero, at least one of those individual parts must be equal to zero.  (because the only way to get zero by multiplying is to multiply by zero.  sorry, you probably already knew all that.  anyway, you get\n\nx=0\nand\nx-1=0,   which you solve to get x=1\n\nthat tells you that the function is undefined at x=0 and 1, but those arent necessarily both VAs.  if any part of the denominator can 'cancel' with the a part of the numerator, its considered a 'removable discontinuity' because it cancels out in theory, but you cant really do that because that would totally change the function.\n\nlong story short, an x cancels out, so the discontinuity you got from that piece (x=0) is a hole in the graph, not a VA.  the x-1 cant cancel, so you know that x=1 is your one and only VA.\n\n\nabout the HA . . .\n\nthe degree of the numerator is actually 2.  remember that the degree is the highest power of x.  since you have x^2 in the numerator, its degree is two.\n\nto get the degree of the denominator, you have to look at what the highest power of x would be if you distributed and FOILed it all the way.  if it helps you, go ahead and do that.  all you really have to do is count the x's that will be multiplied together though.  there are 2, which you know will give you x^2, so the degree of the denominator is also 2.\n\nthe rule is that since the degrees of the numerator and denominator are the same, your HA will be y=(leading coeffiecient of the numerator)/(leading coeff. of the denominator)\n\nremember that the leading coeff. is the coeff of the x with the highest power.\n\nfor the numerator thats easy, its 1\n\nfor the denominator, you have you use the distributed version, which is also 1.\n\nfollow the rule, and y=1/1, or y=1\n\n\n\nthere you have it!  final answer . . .\n\nVA: x=1\nHA: y=1\n\n\nof course feel free to send me a message if you have a question :)
605867
(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c
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.
867867
Ignore the relative maximum from -2<x<3. The horizontal asymtote is defined to be the limit as x approaches infinity of the function. Which as you have calculated is 3.
820698
Where does tan x have vertical asymptotes?\nAt -pi/2, pi/2, 3*pi/2, 5*pi/2 etc.\n\nSo this graph you are dealing with must have asymptotes where\n2x - pi/4 = -pi/2, pi/2, 3*pi/2 etc.\n\nAdd pi/4 and divide by 2, and you should find that only two of the results are within the domain (0 to 2pi) you've been given, so they are the asymptotes you want.
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)
280036
The vertical asymptotes are (generally) those values that x cannot be.  For example with this function:\n\ny = 1/(x(x-1))\n\nx cannot be 0 or 1, because that would mean dividing by zero.  Thus this function has vertical asymptotes at 0 and 1.\n\nFor horizontal asymptotes, you're looking for values that y can never be.  Again:\n\ny = 1/x\n\nhas a horizontal asymptote at 0, because no matter how large x gets y will never be 0.  In this case look for values of y when x approaches infinity.\n\nOf course there are more complicated types of asymptotes, but this is a good introduction.