Indentify all the vertical asymptotes for 0< or equal to x <2pi, y =tan(2x-(pi/4))?
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.
906157
foci: (h, k ± c) h = 0, k = 0, c = 8  (0, 0 ± 8)\n\nasymptotes: y - k = ± (a / b)(x - h)\ny - 0 = ±(a / b)(x - 0)\ny = ±(a / b)x\na / b = 4\n\nc^2 = (4b)^2 + b^2\n64 = 16b^2 + b^2 = 17b^2\nb = √(64 / 17) = 1.94\n\na = 4b = 4(1.94) = 7.76\n\n(x^2 / 60.24) - (y^2 / 3.76) = 1
617377
first put it into the form (x-h)^2=4p(y-k)\nCenter: (h,k)\nFocus: (h,k+p)\nFocal Width: 4p\n\nPlot those points and you should be good!\n(if you're having trouble getting it into that form try using the complete the square method)
346140
Well there are the obvious values that are known of arctan. But if your constant K doesn't fall into these categories I'm not sure how you would do this without taylor expansion.\n\nTry this link, it gives the well known values on tan(x) so reading it backwards gives the arctan values
697906
let y = arc tan (x)\n\ntan y = x\ndifferentiate\n\nsec^2 y dy/dx = 1\n\nor(1+tan ^2y) dy/dx = 1\n\ndy/dx = 1/(1+tan^2 y) = 1/(1+x^2)
970213
Use the Chain Rule repeatedly:\n\nOriginal function: y = (tan(sin x))^2\n\ndy/dx = 2 tan(sin x) (d/dx) tan(sin x)\n\n= 2 tan(sin x) (sec(sin x))^2 (d/dx) sin x\n\n= 2 tan(sin x) (sec(sin x))^2 cos x, QED
885970
Domain: All real numbers\nNo vertical asymptotes\nHorizontal asymptote: lim_x->inf 1/(x+9/x) = 0\nCritical points:\ny' = [x^2+9-x(2x)]/(x^2+9)^2\n= [9-x^2]/(x^2+9)^2\nx = -3, 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
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
853022
It should be a function because one x input value gives only one y output value. Try the vertical-line test. But the function does not have an inverse though.
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.
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.
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
6838
∫xSec(x²)²dx = .5Sin(x²)/Cos(x²)
890950
Did they tell you that\nIf x is very close to 0, then sin(x)=tan(x)=x;\n(sin(x/2))^2 / x^2 = (x/2)^2/x^2=1/4\n(tan(2x))^3 / x^3 = (2x)^3/x^3=8
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.
1113206
0, pi/2, pi, 3pi/2, 2pi\n\nall values where 2x = a muliple of pi