College Algebra function problem. Please Help!!!!?
831482
We just started this stuff in school.  first, think of f(x) as y.  then make a table with the left column as x, and the right column as y.  the vertex is going to be -b/2a.  so in this case.  it'll be -1/4.  that's the middle number in the table and also the vertex.  now put above that number in the table on the x column as so.\nx  y\n1\n0\n-1/4\n-2.5\n-3.5\nnow fill in the the x column by plugging in the x variables.  plug in 0 for x and thats your y-intercept.  then plug in your 0 for y, and thats your x-intercept.  this is going to form a parabola so the axis of symmetry is -1/4.  the domain is going to be all real numbers because there are no contradictory numbers to fill in.  Im not sure how to do the range because im only a junior in high school but I hope I helped a little.
940363
About Me\nMember since: October 03, 2006\nTotal points: 100 (Level 1)\nPoints earned this week:\nTotal answers:\nBest answers:\n\nI need math help again... PLEASE HELP! I m desperately in need.?\n\nAnswer PARTB and PART C in the space provided. Evaluation of your solution to each problem will be based on the following.\n\n*A correct mathematical method for solving the problem is shown\n\n*The final answer is accurate and a check of the answer is shown where asked for by the question.\n\nDetermine if the equation is a linear equation, and if it is, write in standard form.\n\na. 5x+1=3y\n\nb. y=2x   x=y/2\n\nc. 4(2y-x)=1  8y-4x=1\n\nd. 25x-2y+2=0\n\ne. 2x-5y/2=7x\n\n2. a. f(x)= x (exponent 2)+2x+1 f(0),f(2)\n\nb. f(x)=2x-5 f(3),f(-3)\n\nc. f(x)=(-3x)exponent 2 f(0),f(-1)\n\nd. f(x)=2x(exponent 3)+4 f(0),f(-1)\n\ne. f={(1,5),(3,7),(5,9),(7,11)} f(3),F(7)
889441
Hello dear Neil \n\n5x^2+3x+1=0\n\n● Step 1 \nif F(x) = ax^2 + bx + c \n∆ = b^2 - 4(a)(c) & ∆ ≥ 0\nF(x) = 5x^2+3x+1\na = +5 , b = +3 , c = +1\n∆ = (3^2) - 4 (5)(1) \n∆ = 9 - 20 \n∆ = -11 < 0\nwell this function has so real solution\n\n● Step 2 \nx1 = 1/10 (-3 - i √11)\nx2 = 1/10 (-3 + i √11)\nso \nx 1 = -0.3 - 0.331662i\nx 2 = -0.3 + 0.331662i\n\nSo you are pretty right .\nGood Luck Neil ♣
104175
f=thanx\ng=for\nx=the\ny=two\nz=points\n\nTell your teacher you need some help with your algebra.
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!!
491024
y = f(x) = 3/4x + 11/4
865391
I think there is a sign error: the cubic term is supposed to be -2x^3, not 2x^3. Then the other roots are 3 and -1 and the factors are\n\n(x-8i)(x+8i)(x-3)(x+1)
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
918705
Don't you see the pattern?\nIt's right in front of your eyes.  Let g(x) be what is in the parentheses.  So g(x) = x+2\nNow replace x+2 with just x, and that will be f(x).  \nSo f(x) = x^3 - 5x^2 + 3x - 1\nAgain, this is assuming that h(x) = f(g(x)) (based on your other problems)
30225
These problems are talking about composite functions, which is where the results of one function are input into the second.  Usually it's written with a small circle in between (I'll use gof(x)) and read "g of f of x".  Remember that the function closest to the variable is the one that acts on it first.\n\nOne helpful tool to use when you can't figure out a problem is simplifying it to a problem you've done before.  For example, if f(x)= 3x and g(x)=x^2 ("x squared") then the composition gof(x) means take the results of f(x) and put them into g(x).  \nSo then gof(x) = (3x)^2 = 9x^2.\n\nYour problem is a bit more complicated, but we can use the same process:  \n\ngof(x) would be g(ax-1) so \ngof(x) = (ax-1)^2 + 3(ax-1) + 1\nUse the distributive property to expand this to\ngof(x) = a^2x^2 - 2ax + 1 + 3ax - 3 + 1\nand combine like terms to get\ngof(x) = a^2x^2 + ax - 1   *\n\nSo far, we've done everything just from the definitions of g(x) and f(x).  Now we use the last piece of information we were given: \nthat gof(x) = 4x^2 + ax + b  *\nWe compare the coefficients on like terms of these two polynomials (with the stars) to determine that since a^2 must equal 4, \na = 2 and b = -1 (from comparison of the constant terms).\n\n\nSee if you can use a similar process for the second problem!
470404
((a^2 - b^2)/8a) X ((4a^3)/(a + b)^2))\n(4a^3(a^2 - b^2))/(8a(a + b)^2)\n((4a^3)/(8a)) * ((a^2 - b^2)/(a + b)^2)\n((a^(3 - 1))/2) * (((a - b)(a + b))/((a + b)(a + b)))\n((a^2)/2) * ((a - b)/(a + b))\n\nANS : ((a^2)(a - b))/(2(a + b))\n\nOn your problem, here is where you went wrong\n\nTo make it easier, you should had kept your problem in\n\n(4a^3(a - b))/(8a(a + b)) form, then had done this\n\n((4a^3)/(8a)) * ((a - b)/(a + b))\n\n(4/8)(a^(3 - 1))((a - b)/(a + b))\n(1/2)(a^2)((a - b)/(a + b))\n((a^2)(a - b))/(a + b)
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
919338
Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.\n\n#1: Apply the product and chain rules to find f'(x), then evaluate at 0:\n\nf'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nNow evaluating :\n\nf'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2\n\n#2: First, evaluate the function at y=1 to get the value of x:\n\nxy^2 -2y + 4y^3 = 6\nx - 1 + 4 = 6\nx = 3\n\nNow, implicitly differentiate both sides:\n\nd(xy^2 -2y + 4y^3)/dx = d(6)/dx\ny² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0\n\nSolve for dy/dx:\n\ny² + (2xy - 2 + 12y²) dy/dx = 0\n(2xy - 2 + 12y²) dy/dx = -y²\ndy/dx = -y²/(2xy - 2 + 12y²)\n\nNow evaluate at (3, 1):\n\ndy/dx = -1/(6 - 2 + 12) = -1/16\n\n#3: First, find the derivative:\n\nf'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)\n\nSimplifying:\n\nf'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²\nf'(x) = (-x²-2)/(x²+x-2)²\n\nNow we must find where f'(x) < 0\n\n(-x²-2)/(x²+x-2)² < 0\n-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)\n-x² < 2\n\nNow, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:\n\n(x²+x-2)²≠0\n(x-1)²(x+2)²≠0\nx≠1 and x≠-2\n\nWhich makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:\n\n(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)
982041
Well dear\n\n-If the first root is " α " and the second roots is " α + 3"  so have  \n▪x ² - (α + β) + α β = 0 \n▪( x - α )(x - β ) = 0 \n-b/a = α+ β  \n&\nc/a = α β\nso we have ; \nc/a = m/4  \n-b/a = - ( -16) / 4 = 4 \nα + α + 3 = 4 , 2α + 3 = 4 \nα = 1/2  ,  β = 3 1/2 \n2 /a = α β  =>  m/4 =  1/2 * 7/2  , m = 7\n\nGood Luck dear
857286
Im not sure don't right this at your Homework.........\nwhere f(x) > 0. a. (–3, 1) b . (–3, 0) ∪ (1, ∞) c . (1, 3) d . (0, 1) ∪ (3, ∞) If f(x) = x(x – 1)(x – 4)2, use interval ..............??????\n\n\nSORRY I DON'T HELP TOO MUCH I TRY MY BEST......
935264
Actually, the answer is *not* 4+x.\n(f/g)(x) = (16-x^2)/(4-x)\nNow, factorise the top:\n(4+x)(4-x)/(4-x).\nSo *if x is not 4* then (f/g)(x) = 4+x. However, the function is not defined at x = 4.
1039099
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.