rational function algebra help?
966227
No oblique assymptotes, since the highest degree in the numerator is the same as in the denominator.\n\nlet's factor both equations:\n\n2x^2-x-10 = (2x-5)(x+2)\n-x^2+2x+24 = (6-x)(x+4)\n\nNow that we have that done, it will be easier.  There are no zeroes in the numerator that the denominator shares, which means that all zeroes of the denominator will be vertical assymptotes.\n\nx=6\nx=-4\n\nFor horizontal assymptotes, multiple the top and bottom by 1/x^2 (highest degree of x).  We get\n\n(2-1/x-10/x^2)/(-1+2/x+24/x^2) As x approaches infinity, this approaches 2/-1 = -2\n\nThus our horizontal assymptote is y=-2
865303
1) If h(x) = f(g(x)), then g(x) = 4+3x and f(x) = x^5\n2) Assuming your function is 3sqrt(x^2 -8), then\ng(x) = x^2 - 8 and f(x) = 3sqrt(x)
819185
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut35_polyfun.htm\n\nhttp://www.sosmath.com/algebra/algebra.html\n\nI found these to be helpful.
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.
779926
Just expand the brackets\n\nf(x)=(3x^3+3x)(x+7)\n\nf(x)=3x^4+21x^3+3x^2+21x\n\nThen differentiate each single term by multiple the coefficient by the power and then decreasing the power by 1.\n\nf'(x)=12x^3+63x^2+6x+21
190708
you've asked a lot of questions regarding rational and irrational. Here's a site that may be of help:\n\nhttp://www.purplemath.com/modules/numtypes.htm
708317
Ok, so if I have this correctly ... f(x) = 3x + 7 and you want to figure out the value of 2f(x²) - 5.\n\nSo first, you have to substitute x² into your equation for f(x) to obtain:\n3x² + 7.\n\nNow, this is the value for f(x²), so we can substitute it into the sexond equation:\n\n2f(x²) - 5\n=2(3x² + 7) - 5\n\nNow just expand and simplify.\n\n2(3x² + 7) - 5\n=6x² + 14 - 5\n=6x² + 9\n= 3(2x² + 3)
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
38332
(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
38209
actually polynomial functions make me happy...they are actually pretty easy.  You evaluate each term (thats the stuff separated by the addition (+) or subtraction (-) sign) individually.  So if you had sin(x) -x^3 + 3x\n\nthen you have the following\n\ncos(x) -3x^2 +3\n\nhope this helps!!
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
52920
The common denominator of this rational equation is 4(x-1)(x+2).  Multiply each term of the equation by this, and all the denominators will cancel, leaving you with a quadratic equation I assume you can solve.\n\nThe easy way to find a common denominator is to just write down all the denominators of the equation.  That's not always the least common denominator, but it is in this case.
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.
110508
(x^2+5) / (x^3+15-2x^2+1)^2\non [0,1]
735529
f(x)=-2x^4+6x+7\nf(-x)=-2(-x)^4+6(-x)+7\n=-2x^4-6x+7\nnether odd nor even
192120
Use the chain rule. Let f(y)=y^(1/2) and g(x)=x-2. Then your expression is just f(g(x)), and the chain rule states that D(f(g(x))) = Df(g(x))*Dg(x) = [(1/2)*(g(x))^(-1/2)] * [g'(x)] = [(1/2)*(x-2)^(-1/2)] * [1].\nI suggest doing several more of these questions until you no longer need to explicitly write out f and g, but can do so in your head.
3037
This is not presented as an equation, so the goal appears to be to combine the rational expression --> x+10 \nThat's all you can do with the question as it is written.