Quadratic Functions question.?
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analyzemath.com/quadraticequations
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1) If h(x) = f(g(x)) = (x^3 -1)/(x^3 +1), then \nf(x) = (x-1)/(x+1) and g(x) = x^3\n2) If h(x) = f(g(x)) = 3sqrt(x^2 -8), then\nf(x) = 3sqrt(x) and g(x) = x^2 -8
919864
1.  ax + xb\n2.  21x^3 + 24x*2y\n3.  24x^2 + 31x + 10\n4.  24x^2 - 8x - 6\n5.  40x^3 + 66x^2 + 38x + 6\n6.  x^2 - 1\n7.  x^3 - 8
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
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By dividing the equation by 2:\n\nx^2 + 3x +1 = 0\n\nThe quadratic formula is = [-b +/- sqrt(b^2 - 4*a*c)]/2a\n\nWhere a =1;     b = 3;       c = 1.\n\nNow, we plug in this values in the quadratic formula, and we'll obtain a value of x, using the minus sign, and other, using the plus sign, as follows:\n\nx1= [ - 3 + sqrt( 3^2  - 4*1*1)]/(2*1)\nx1= [ - 3 + sqrt(9 - 4)]/2\nx1= [ - 3 + sqrt(5)]/2\nx1= [ -3 + 2.236]/2\nx1= -0.382\n\nx2= [ - 3 - sqrt( 3^2  - 4*1*1)]/(2*1)\nx2= [ - 3 - sqrt(9 - 4)]/2\nx2= [ - 3 - sqrt(5)]/2\nx2= [ -3 - 2.236]/2\nx2= -2.618
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Tell your teacher this is a poorly worded question. ALL quadratic functions have two solutions. You could have two real solutions, one real solution that is repeated once, or two imaginary (complex) solutions.
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b^2=ac\nb^2x^2=ax*cx\n(bx)*(bx)=ax*cx\nflipping\n1/(bx)*1/bx=1/ax*1/cx\ntaking log\nlog1/bx+log1/bx=1/logax+1/logcx\n2/logbx=1/ax+1/cx
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it doesn't factor out...are you sure you typed it correctly?\n\nif it is correct, then use the quadratic formula to get the answer...but its pretty complicated (negative b plus or minus the square root of b squared minus 4ac)...
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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]
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4 (x^2 + 3X -4)\n4(x+4)(X-1)
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x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
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f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
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1.f(x)=x^2+4x-7\nx=[-4+/-rt(16+28)]/2\n=[-4+/-2rt(11)]/2\n=-2+/-rt11\n\n2.3x^2-11x-20=0\n=3x^2-15x+4x-20=0\n=(3x+4)(x-5)=0\nx=-4/3 or 5
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f(x) = x^2 + 7x\n\nf(q+2) = (q+2)^2 + 7(q+2)\n\nf(q+2) = q^2 + 4q + 4 + 7q + 14\n\nf(q+2) = q^2 + 11q + 18\n\nf(q+2) = (q + 9)(q + 2)
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x^2-5x-2=0\n\nuse the quadratic formula\n\n[5  +/- sqrt(25-4*1*-2)]/2\n\n[5  +/-   sqrt(25+ 8)]/2\n\nI distributed the 1/2...easier to type\n\n5/2  +/-sqrt33/2
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This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
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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 ♣