i need help with my algebra2 hw finding vertical asymptotes please if u can help say so?
411240
you get vertical asymptote when denominator of a fraction hits zero. so \nx=0 for 3/2x\nx=-1 and x=3 for (x/(x+1)(x-3))\nand so on
559279
Correct!\nOnly the second x's in the equation for asymptotes will be y.\nThis is an interesting result. Rectangular hyperbola is the locus of far corner of all rectangles' vertices having a fixed area A. And having the asymptotes as two adjascent sides of the rectangle.\nWith proper substitution, the general eqn can be transformed to ione.
774094
- Find the x-intercepts, by setting y=0 and solving the polynomial equation. You may need to factorise\n\n- Find the y-intercepts, by setting x=0 and solving.\n\n- Find stationary points by differentiating the polynomial equation wrt x and setting the differential to 0. These give you the stationary points.\n\n- Differentiate again to find out whether these turning points are maxima, minima or points of inflexion:\nIf d²y/dx²  is positive, then it is a minimum point.  \nIf d²y/dx² is negative, then it is a maximum point  \nIf d²y/dx² is zero, then it could be a max, a min or a point of inflexion.\n\n- Find asymptotes. You may get a horizontal asymptote when x approaches +ve or -ve infinity and a vertical one as y approaches +ve,-ve infinity.\n\nThis is the basic info you need to sketch a graph depending on how complex the polynomial is.
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.
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.
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
1073788
points of inflection are the places where the second derivative of a function equals zero. \nSo in this case f'(x)=6x^2-10, and f''(x)=12x. Now if you set f''(x)=0 it is the same as 12x=0 and dividing both sides by 12 you have that x=0. Thus your inflection point is at x=0.\n\nNow recall that f(x)=2x^3 -10x +3 is a third degree polynomial, which looks some what like and s curve, and it does indeed have only one place where the concavity changes from up to down, so having only one inflection point makes sense.\n\nPolynomials do not have vertical asymptotes nor do they have horizontal ones. You might be thinking about rational functions, which are polynomials divided by polynomials.\n\nIf that is indeed what you are refering to them, to find the vercital asymptotes you set the denominator of the function equal to zero and solve for x. These are then the vertical asymptotes. \n\nFor horizontal asymptotes it is more complicated I suggest you look it up in a math book, the simplst case is like this:\n\nIf the highest degrees on the numerator and the denoinator are equal then the ratio of the coefficients is the horizontal asymptote. Example f(x)=(2x^3 -10x +3)/(3x^3-17), the highest degree of the umerator is 3 which is the same as the highest degree of the denominator, so the horizontal asymptote is y=2/3.\n\nHope this helps!
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^_^
317924
See previous answer--\n\nGives vertical height as rt27\n\nTherefore 1/2*6*rt27 is area of one side\n\nBase is 6*6 = 36
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.
1126492
Use pythagorean theorem.\n\n3^2 + 7^2 = hypotenuse (diagonal) squared\n\n9 + 49 = x^2\n\n58 = x^2\n\nsquare root of both sides  diagonal is = 7.61577  feet\n\n(or 7 feet 7.4 inches)
803546
Use the pythagorean Therym(?)\nThat is, well, here is an example:\nIf one leg of a right triangle is 6 units, and the other leg is 8 units, what is the legnth of the hypotenuse?\nthat is:\nleg 1=a, leg 2=b, and the hypotenuse = c\nThe hypotenuse always = c, it is the side parallel to the right angle of the triangle.\nthe pythagorean therym(?) is a^2(a squared)+b^2(b squared)=c^2(c squared)\na=6, b=8, and c=?\n6^2+8^2=c^2\n36 + 64= c^2\n100= c^2\nc=10^2\nso the missing side would b 10 units, just use that example for your problem, and remember, if you are trying to find the legnth of a leg, just plug in the one leg length that you have (this would be "a") and the hypotenuse(this would be "c") and subtract the square of "a" from "c". find the square root, and vua la, there you have it. hope that helps
555107
MY guess is that your trying to find the Hypotenuse of a triangle.  Thus A^2+B^2=C^2 or in this case H^2+W^2 = C^2, 25 + 196 = C^2, 221 = C^2, C = 14.86.  Missing side is 14.86.
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 :)
1127430
try this one http://www.algebrahelp.com/
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.
192101
The vertical asymptote of (1/x+1) is at x = -1 because when x = -1 you have (1/(-1+1)) = (1/0) and you cannot divide by 0.