What are all the vertical asymptotes for 0< or equal to x < 2pi, y=tan(2x-(pi/4))?
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
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.
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
694637
tan(theta)=x/x^3=1/x^2. So theta=arctan(1/x^2)
1050478
tan(x) = y  implies arctan(y) = x\n\narctan ( x )  = tan-1 ( x )\n\ndoes this help u
951523
vertical: x=3\nhorizontal: y=5\n`
791450
assuming h is a real number, we have:\nto find roots of any function set it equal to zero\n 1 - tan(h)*[2*pi*h/x]^2 = 0\n      tan(h)*[2*pi*h/x]^2 = 1 \ntan(h)*[4*pi^2*h^2/x^2] = 1\ntan(h)*[4*pi^2*h^2] = x^2  x ~= 0 (not equal to zero)\nx = +/- sqrt(tan(h))*2*pi*h these are the roots
653191
You have to look for the vertical asymptopes. For exapmle in (x^2-5)/((x+2)(x-5)), the vertical asymptopes are the ones that will make the denominator 0, so -2 and 5. With these, as the function approaches them, it will rise to infinity, or unless you have a horizontal or slant asymptope.
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.
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.
901871
It's 9.\ntan(beta + pi) = tan(beta + 2pi) = tan(beta) = 3\n\nSo tan beta + tan(beta + pi) + tan(beta + 2pi) = 3+3+3 = 9.\n\nYou can see why tan(beta) = tan(beta + pi) = tan(beta + 2pi) by looking at this graph of y = tan x\nhttp://www.intmath.com/TrigGrph/4_TCSC.php\nYou can see that the graph repeats itself in intervals of pi. So if you look at where y = 3, then move pi spaces to the right on the x-axis, you will see that y=3 there as well.
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.
953056
No asymptote.  It just keeps growing. That is, as x─►±∞, y ─►±∞.
11152
Here is one, although rather inefficient.\n\narctan(1) = pi/4, so pi = 4arctan(1)\n\nThe Taylor series for arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + ...\n\nso pi = 4arctan(1) = 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - ...)\n\nTo get N decimal place accuracy,  the last term you add in must be less than .5 x 1/(10^N).\n\nThere other fuctions you can find that will converge more quickly with fewer iterations, but this will work.
782645
1/x=y+3\nit resembles the equation of the asymptotes of hyperbola
921171
If tan( t ) = -2 then t = atan( -2 ) = -1.107 + multiples of π.\n\nIn the range from 0 to 2π the required angles are\n\n-1.107 + π and -1.107 + 2π, or\n\n2.034 and 5.176