Find all horizontal and vertical asymptotes (if any)?
827399
V.A. : none\nH.A. : y=3 (  because  if the degree of the numerator x^2 is equal to the degree of the denominator ,you should divide the coefficient of the numerator by the coefficient of the denominator,so 3/1 = 3)
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
1114168
Ok, here's how i think you do it.\n\nto get the horizontal asymptote, you have to get the limit of the functon when x tends to be infinite, so you would get something like this\n\nlim     (3x-x^2)/(4-x)\nx→∞\n\nthen you substitute ∞ on the x's and you get\n\n∞/∞, that's not defined, so what you use is l'hopital law where you derive both the nominator and the denominator of a fraction, so you will get\n\nlim    3-2x/-1\nx→∞\n\nnow you can substitute ∞ and you get\n\n(3-2*∞)/-1 =∞\n\nand you do the same with -∞ and you get as an answer -∞\n\nwhich means there is no horizontal asymptote.\n\nI hope this helps.
769990
vroma.org/~bmcmanus/\nclothing.html
1050977
http://mscir.tripod.com/parabola/
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
357173
This should help.\n\nhttp://www.geomatters.com/products/category.asp?CID=1
910564
umm none that im aware of, what kind of mathamatics leval are you looking at?
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)
742119
asymptote at x=5 and y=2
205031
There will be a vertical asymptote at x=-2, assuming you meant to put "x+2" in parenthesis. When x values get increasingly closer to negaitve two from the right, the graph approaches infinity. When the x values approach -2 from the left, the graph approaches negative infinity.\n\nSimilarly, there is also a horizontal asymptote at y=3. As x approaches infinity, the graph approaches 3 (1/infinity + 3 = 3). Also, as x approaches negative infinity, the graph also approaches 3 (1/negative infinitive + 3 = 3).
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.
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
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.
218516
If that is supposed to be a function of x, then\nf(x) = (-2x + 7)/(4x - 3)\nSince the denominator cannot be zero, equate it with zero and you get the vertical asymptote:\n4x - 3 = 0\n4x = 3\nx = 3/4 --- Vertical asymptote.\n\nTo get the horizontal asymptote, get the inverse function.\ny = (-2x + 7)/(4x - 3)\n4xy - 3y = -2x + 7\n4xy + 2x = 3y + 7\nx = (3y + 7)/(4y + 2)\n\nThen the horizontal asymptote is the denominator equated to zero.\n4y + 2 = 0\n4y = -2\ny = -1/2 --- horizontal asymptote\n\n^_^
367516
here are the anwsers in order from top to bottom \n-T\n-T\n-F\n-F \nhope this helps