How do you solve this asymptote question?
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
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
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
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
707497
Horizontal aymptote? In x^3 + y^3 = 3xy as x gets big y gets big too, there is no horizontal asymptote.\n\nTo get an idea of the asymptote, look at x getting really big. In this case the xy term is much smaller, and so you ignore it, so you get x^3 = -y^3, x = -y as an asymptote.
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.
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.
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
834638
1. horizontal asymptotes : find the limit of fn as x--> infinity:\n\ndivide the fn's num and denom by highest i.e. x^2:\nr(x) = {x + (3)}/{1  - (4/x^2)}\nhence as x--> +-inf, r(x)---> +- inf\n\n2. vertical asymptote means find zeroes of denom:\nx^2-4=0\n(x+2)(x-2)=0\nhence x= 2 and x=-2 are the answers.
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
845874
1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
780025
x ≤ 5\n\n-- - - - - - -s-
266735
Consider the equation y= a (x^n) / b (x^m) where n and m are the highest powers of the variable x found AFTER factoring and reducing the numerator and denominator to the products of the smallest polynomials posible. The situation may be that n or m may be zero, and that's OK.\nThe y intercept is found when x=0. The x intercept is found when the factors of the numerator are set = to zero; i.e. if y= x^2 + 3x + 2 = (x+2) (x+1), then x=-2 and x=-1 are conditions that yield y=0, and -2 and -1 are the roots of the numerator and are the y intercepts.\nThe vertical asymptotes are when Y becomes infinite, i.e the denominator becomes zero. Take each of the factors in the denominator, set them equal to zero, and solve for the corresponding values of x.  For an equation y = 6 / (x-2) (x+3), set each denominator factor equal to zero, and the conditions are for two vertical asymptotes, one at x=2, and the other at x= -3. \nFor the horizontal asymptotes go back to ,  y= (a (x^n)) / (b (x^m)). If n>m there is no HA. If n<m, such as y= x / ((x^2)+1), the HA is the x axis. If n=m the HA is an/bn, or just a/b.\nTry this. I hope I didn't make any typo mistakes
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
776830
X= (2k + 1)/n * (pi/2)\n\nThis tells you the vertical asymptotes for y = tan (nx)\n\nk is any integer\neg the graph of y = tan 3x has vertials asymptotes at:\n\nx = π/2, ±3/3 π/2, ±5/3 π/2, ±7/3 π/2, .....\n\n2k + 1 = ±1, ±3, ±5, ±7, ....... and n = 3
933488
As x becomes infinitely large in either the positive or negative direction, f(x) approaches the line y = 2x because the constant terms are negligible compared to the x^3 and x^2 terms, so the ratio 2x^3 / x^2 = 2x dominates.  There is no horizontal asymptote, but there are vertical asymptotes at x = 4 and x = -4 because the denominator has roots there.
764329
x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.