is parabola(rational funtion) a polynomial?
931584
Is parabola(rational funtion) a polynomial?\n\nok, you know that a parabola is not a rational function. a parabola is the graph of a quadratic function that looks like:\n\nf(x) = y = ax^2 + bx + c. \n\na polynomial is a sum of finite monomials in which coefficients of terms are real numbers and exponents of variables are whole numbers.\n\nthus, a quadratic is a polynomial.\n============================================\n\n\nnow, is hyperbola a polynomial?\n\nwhat does a hyperbolic function look like?\n\n(x-h)^2/a^2 - (y-k)^2/b^2 = 1\n\nif you solve for y, we do not have a polynomial.
888635
Let f(x) = 3x^4 - 25 + 5/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nBecause f(x) is a polynomial function (of degree 4) on x, it follows that the function f is continuous on R (the set of real numbers).  So, if we take the limit on the left (and right) of x, we will get the same value for both sides. \n\nThat is, the behavior on the left (Left-hand Limit) will be: as x approaches  c from the left, we get f(x) approaches f(c) = 3c^4 - 25 + 5/4. Similarly, the behavior on the right (Right-hand Limit) will be: as x approaches c from the right, we also get f(x) approaches f(c) = 3c^4 - 25 + 5/4. \n\nTherefore, both the left and right-hand limit yields to f(c), i.e. as x approaches c, we always obtain f(c).\n\nHence, both left and right-hand behavior of f are the same for all real valued x.
708917
(x^4) + (4x^3) + (2x^2) - 8x - 8=0\n\nGeneral:  ax^4+bx^3+cx^2+dx+e=0\nif p/q is a rational root then p is divisor in a and q is divisor in e =>\n\nin this case:\np belong to {-1,+1} & q belongs to {-8,-4,-2,-1,1,2,4,8} =>\n\np/q belong to {-8,-4,-2,-1,1,2,4,8}\n\nWe try  -2 as a root and see\n2^4 + 4*2^3 + 2 * 2^2 - 8 * 2 - 8=0\n\nWe can now do polynomial division by (x+2):\n\n(x^4) + (4x^3) + (2x^2) - 8x - 8 : (x+2) = x^3 +2x^2 -2x - 4\nx^4 . . . + 2x^3\n- - - - - - - - - -\n0 . . . . . . 2x^3\n. . . . . . . .2x^3 . .+4x^2\n- - - - - - - - - - - - - - - \n. . . . . . . . .0. . . -2x^2 . . -8x\n. . . . . . . . .  . . . -2x^2 . . -4x\n- - - - - - - - - - - - - - - - - - - -\n. . . . . . . . . . . . . . . . . . .-4x. -8\n. . . .  . . . . . . . . . . . . . . -4x  -8\n- - - - - - - - - - - - - - - - - - - - - - -\n. . . . . . . . . . . . . . . . . . . . . .  0\n\nexactly the same procedure as before gives x= -2 is root so now we go again\n\nx^3 +2x^2 -2x - 4 : (x+2) = x^2-2\nx3^. .2x^2\n- - - - - - -  - -\n. . . . . . .0 . .-2x -4\n. . . . . . .0 .  -2x -4\n- - - - - - - - - - - - -\n. . . . . . . . . . . . .0\n\nthe last term x^2-2=0 => x=±√ 2\n\nnow we have :\n\n(x^4) + (4x^3) + (2x^2) - 8x - 8 =\n(x+2)^2 * (x-√ 2) * (x+√ 2) = 0\n=======================\nso x = { x in R I x = -2 or x= -√ 2 or x= +√ 2}\nand x= -2 is root 2 time\n=================================
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
670280
If you have a parabola in the form of y = a(x - h)^2 + k, then you can read off a lot of your particulars directly from the equation:\n\nvertex: (h, k)\ndirection: for a>0, parabola opens up, for a<0, parabola opens down\naxis of symmetry: x = h\n\nSo you have a parabola that opens down with vertex at (-2, 3), and axis of symmetry at x = -2.\n\nNow compare this to y = x^2: opens up, vertex is at (0, 0), and axis of symmetry at y = 0.  Your first equation will also be skinnier because of the a value: if |a|<1, your parabola is wider than y = x^2, and if |a|>1, then your parabola is skinnier than y = x^2.
83507
If you plot the two parabolae very carefully, you will see that they do not intersect, at least in real space. However, if you allow complex roots to the resulting quartic equations, then one can of course find 4 complex roots. \n\nso y takes the values:\n\n-6.0583   +   i*0.5986\n-1.9416  +   i*0.3398\n-6.0583   -   i*0.5986\n-1.9416  -  i*0.3398\n\nand at each of those values of y, x takes the corresponding values:\n\n   3.8783 - 2.4642i\n   4.1215 + 1.3989i\n   3.8783 + 2.4642i\n   4.1215 - 1.3989i
208169
Factoring, I guess, but it really depends on the type of polynomial question. Can you give a sample?
423853
[2x^{2}+ 2x2y]+[2x2y+2x^{2}]+\n[6yz^{9}+3y^{2}9z^{8}]+\n[3y9z^{8}+y^{3}72z^{7}]\n\nthis is using z as the last term.  Based on what I used, I think a 3D parabola of sorts will emerge
799253
Polynomial is a general term for any expression, including monomial, binomial, trinomial, etc.  Monomials have just one term with no + or -, such as x or 3x or 5x^6.  Binomials are two monomials added or subtracted, such as 3x + 5 or 5x^2 - 6x.  The degree of a polynomial is the highest exponent of any of its variables, assuming there is only one variable.  So for x^3 + 8x^2 + 3x - 4, the degree is 3.
1154261
The third answer above is close, but not quite right.  A function can have two different horizontal asymptotes -- one in the limit of +infinity, and the other in the limit of -infinity.  You need to check the values of both limits separately.  \n\nFor instance, the horizontal asymptotes of arctan(x) are +pi/2 (as x -> +infinity) and -pi/2 (as x -> -infinity).\n\nFor rational functions (functions that can be expressed as the ratio of two polynomials) there is an easy way to test for and find the horizontal asymptotes:\n\nLet  f(x) = p(x)/q(x) be a rational function. Then: \n\n1.    If (the degree of q(x))  > (the degree of p(x)) then there is one horizontal asymptote at 0.\n\n\n2.    If (the degree of q(x)) =  (the degree of p(x)) then there is one horizontal asymptote at the (leading coefficient of p(x))/(leading coefficient of q(x))\n\n\n3.    If (the degree of p(x)) < (the degree of q(x)) then there is no horizontal aymptote.
1037602
A parabola is the set of all points equidistant from a fixed point and a fixed line in the plane.  The fixed line is the directrix.  Since you are wanting a directrix y = -4, you are wanting a parabola that opens upward.  \n\nThe general form for a parabola that opens upward is:\n\n(x-h)^2 = 4p(y-k)\n\nwhere (h,k) is the vertex and y = -p is the directrix.\n\nSince you want a vertex of (0, 0), this means that your h and k are both 0.  The p in question is always greater than zero.  You have y = -4, so p = 4.  \n\nTherefore, we have (x-0)^2 = 4*4*(y-0)\n\nSo our equation is x^2 = 16y.\n\nAs far as the the standard form goes, different books refer to various forms as "standard."  The most common standard form for a parabola of this nature is to solve for y.\n\nSo we get:  y = (1/16)x^2\n\nAngel
891987
The first step is to apply the Rational Roots Test to the polynomial to get a list of values that might possibly be solutions to the polynomial equation. You can follow this up with Descartes' Rule of Signs, if you like, to narrow down which possible zeroes might be best to check. Of course, if you've got a graphing calculator, it's a good idea to do a quick graph, since x-intercepts of the graph are the same as zeroes of the equation. Seeing where the graph looks like it crosses the axis can quickly narrow down your list of possible zeroes.\n\nOnce you've found a value you want to test, you use synthetic division to see if you can get a zero remainder. If you get a zero remainder, you've not only found a zero, but you've also reduced your polynomial by one degree. Remember that synthetic division is, among other things, division, so checking if x = a is a solution is the same as dividing out the linear factor x – a. This means that you should not return to the original polynomial for your next computation (for finding the other zeroes); you should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.\n\nYou should expect to get some complicated solutions (that is, solutions containing square roots or complex numbers, or both). You will find these zeroes by applying the Quadratic Formula for the final quadratic factor.\n\nFind all the zeroes of the following polynomial: 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 \nFirst, I'll apply the Rational Roots Test--\n\nWait. Actually, the first thing I'll do is check to see if x = 1 or x = –1 is a root, because these are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because you can just look at the powers and the numbers. When x = 1, the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1 isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so x = –1  is a root, and I can do the synthetic division with x = –1 right now.\n\n\n\nThis leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x + 24. (Since I've divided out the factor x + 1, I've reduced the degree of the polynomial by 1. That's how I know this is a degree-four polynomial.) Now I'll apply the Rational Roots Test to get a list of values to look at:
824215
Did you mean -(3/4) * (x^3+2x^2-3x+1) ?\n\nLet f(x) =  -(3/4) * (x^3+2x^2-3x+1). Equivalently, f(x) = -(3/4)x^3 - (3/2)x^2 + (9/4)x - 3/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nSince f is a polynomial function (of degree 3) on x, it follows that this is continuous on the set of real numbers. Thus, by applying the limit on both sides (left and right) of x, we will obtain the same value for both sides. \n\nThat is, the left-hand Limit will be as follows: as x approaches to the left of c, we make f(x) approaches f(c) = -(3/4)c^3 - (3/2)c^2 + (9/4)c - 3/4. Similarly, the right-hand limit will be: as x approaches to the right c, we also get f(x) approaches to f(c). \n\nTherefore, both the left and right-hand limit yields to f(c); so as x approaches c in any direction, we always obtain f(c). Hence, both left and right-hand behavior of f are the same for any choice of a real valued x.
1078301
Graphs of quadratic equations yield parabolas, and parabolas are symmetrical about its axis of symmetry.\n\nIn completed square form, a parabola is:\ny - k = a(x - h)^2\nThe parabola is symmetrical about x = h\n\nIn standard form, a parabola is:\ny = ax^2 + bx + c\nThe parabola is symmetrical about x = -b/2a
565133
We start by factoring the polynomial.  A little trial and error (testing the factors of 2, which is the constant term) reveals that 2 is a root.  So, we know that (x-2) is a factor.\n\nDividing the polynomial by (x-2) using synthetic division, we find the other factor is 6x^2+x-1.  This can be factored as (3x-1) (2x+1).\n\nSo, the equation, in factored form, is\n(x-2)(3x-1)(2x+1)=0\n\nSo, the roots are x=2, x=1/3, and x=-1/2.
1076159
Hi:\n\nthere are 6 formulas for a parabola :\n\n1) x^2+y^2= e^2(d+x)^2    ( explain what the variables are below\n as long as e=1 it is a parabola \n\n\n2) r = de/(1-e cos(thea))\n\n3) (y-k)^2=4a(x-h)  make a Y parabola\n\n4)  (x-h)^2=4a(y-k) make a X parabola\n\n5)     Ax^2 + bx+ C= y  ( as long as A does not equal zero)\n\n6 )        A( x-b)^2 + c= y\n\n\n x = x corr.,  y= y corr,  k =distance from y axis,  h= distance from x axis,  d= is the distance from focus to directrix, e = eccentricity  r = radius