rational functions question?
926034
If by the whole you mean where the finction has a hole in the graph (a removeable discontinuity), then that is when the denominator equals 0.  so there is a hole at -3.  And, yes, I belive that the roots are both -3.
364914
A rational function is the quotient of two polynomials.  A vertical asymptote only occurs when the denominator = 0.  So, any quotient of two polynomials such that the denominator is not = 0 would answer your question.
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
838016
This is basically just asking what is f(x)/g(x)\n\nso your answer would be \n\n(x+2) / (3x^2 - 5)
873165
f(g(x)) = (g(x)+4)^2 = (8x-8+4)^2 = (8x-4)^2 = 64x^2-64x+16\n\nThat's your composite function.\n\nHope this helps.
953940
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
598723
possible rational factors are p/q where, p is factors of the constant and q is factors of the leading coefficient.\np = 15 and q = 2\nfactors of p: 1, 3, 5, 15\nfactors of q: 1,2\nPossible p/q = (plus or minus) 1, 3, 5, 15, 1/2, 3/2, 5/2, and 15/2\nKeep picking one of these values and test it until you get a remainder of 0.\n1 works\n1 2 3 -11 -9 15\n ___ 2  5  -6 -15\n__2 5 -6 -15 0\nThus,\n2x^4 + 3x^3 - 11x^2 - 9x + 15 \n= (x-1)(2x^3 + 5x^2 - 6x -15)\n2x^3 + 5x^2 - 6x -15 can be factored by grouping.\n(2x^3 + 5x^2) + (- 6x -15)\n= x^2(2x+5) -3(2x+5)\n= (x^2-3)(2x+5)\nAnswer: (x-1)(x^2-3)(2x+5)
986272
f(x) = 3(2/x + 1)\ng(x) = x^2\n\n\nHope this helps.
941571
1) f(x) = A(x-1)(x-2)(x-4) for some A. Substitute in 0 to get -16 = A*-1*-2*-4, so A = 2. So 2(x-1)(x-2)(x-4).\n2) Whenever you have a polynomial with real coefficients and a+bi is a root, you also have a-bi is a root. So you know the 3 roots are 4, 3+i, 3-i. Thus f(x) = (x-4)(x-(3+i))(x-(3-i)). Multiplying those last two together, you get x^2 - x(-3-i-3+i) + (3+i)(3-i) = x^2 - 6x + (9-i^2) = x^2-6x+10. So f(x) = (x-4)(x^2-6x+10).\n3) I don't think you wrote the question down properly. Expanding the left hand side gives 3x^3 + (-a-27)x^2 + (9a+42)x - 14a. You need a=5 for the first part to work, but that doesn't give what you wrote.
762368
NB! sqrt(x)=x^½ \n2*(2+1)^½ / -5*(2+3)^½<=>\n( 2*(3)^½*(5)^½ )/(-5*5)<=>\n( -2*(3*5)^½ )/25<=>\n( -2(15)^½ )/25<=>\n-2( 3*5/5*5*5*5)^½<=>\n-2(3/5*5*5)^½= -sqrt(3/5)*2/5\n=========================
811480
1)  I'm afraid I don't understand the question.  It's possible that  you've been taught a different technique than I teach.  The method I teach is essentially the long division you learned in third/fourth grade, modified to work with polynomials.  I'm afraid I'm no help there.\n\n2)  Those are patterns.  You need to understand them and learn to recognize them.  For instance:\n\nx^2 - y^2\n\nis a difference of squares.  That means it will factor into (x+y)(x-y).\n\n9a^2b^4 - 4x^6\n\nis also a difference of squares.  It will factor into\n\n(3ab^2 - 2x^3)(3ab^2 + 2x^3)\n\nThat's what I mean by being able to recognize them.\n\n3)  The only other thing that's important is that the pattern repeats forever.  i^5 = i, i^6 = -1, and so on.  So, i^400 = 1.\n\nAs for the rest of question 3 - well, it depends on what you're being asked.  If you're asked to find all of the rational solutions, then of course you'd find the rational solutions.  If you're asked to find ALL of the solutions, that would certainly include the rational ones.  If you're asked to find all imaginary or complex solutions, then you wouldn't necessarily look for the rational ones, but it might make it easier to solve the problem if you did.  As a thumbrule, I'd suggest always looking for rational solutions unless you have a specific reason not to.\n\nAs for your additional question:  x^2 + 16 = (x+4)^2 just doesn't work.  Try reversing it:\n\n(x + 4)^2 = (x + 4)(x + 4) = x^2 + 4x + 4x + 16 = x^2 + 8x + 16\n\nYou're missing the cross-terms if you try to factor x^2 + 16 that way.  In fact, a sum of squares does not factor over the real numbers.\n\nIf you're ever unsure why a particular factoring pattern doesn't work, you will almost always get an answer by trying it the other way.
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.
873410
1.  (x + 9)(x - 9)\n2.  (t + 2)(t - 2)\n3.  (b + 12)(b - 12)
695915
Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x