Quadratic Functions questions?
963931
If Yahoo answers utilized graphics or filesharing capabilites, I would've shown you how in a minute...  (Shame on U, Yahoo Answers!!)\n\nCheck out what the other guys have underneath, though...
657050
1) f'(x) = 40x^4+6x^2+4\nTo find local max's and min's, set this equal to zero, but you will see that all solutions will be imaginary, so there are no local max's and min's on the real line.\n\nf''(x) = 160x^3+12x\nTo find the points of inflect, set this equal to zer, but you will see that the only place where there is an inflection is a x = 0.\n\nSo what have we learned?  f(x) is a strictly increasing function, meaning tha there could only be one real root.\n\n2. Let u = 1+t and use substituion.
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
912600
-1 < k < 3\n\nthis monic quadratic is a parabola.\na parabola (in fact any equation) will be entirely positive when it is above the x-axis for all x (since it is monic it will open upwards and so it will be entirely positive when there are no real roots ) \nthere are no real roots when b^2-4ac < 0.\n(-2k)^2 - 4(3+2k) < 0\n4k^2 - 4(3+2k) < 0\nk^2 - 3k - 3 < 0\n(k-3)(k+1) < 0\n-1 < k < 3\n\nI hope this has helped you.  Please remember to choose a best answer for all of your questions.
891271
first into the second:\n\nx^2+(x-1)^2=25\n\nexpand:\n\nx^2+x^2-2x+1=25\n\nget 0 on the right side:\n\n2*x^2 - 2*x -24 = 0\n\na=2, b=-2, c=-24\n\nuse the root finding formula for quadratoic equation:\n\nx=(-b+/-sqrt(b*b-4*a*c))/(2*a)=(2+/-sqrt(4+4*2*24))/4=0.5 +/- 3.5\n\nyour solutions are: x = -3 and 4\n\ncheck:\n(-3)^2+(-4)^2 = 9 + 15 = 24, correct\n(4)^2+(3)^2 = 15 + 9 = 24, correct\n\nsubstitute to the first eqn: y = -4 for the 1st. x and 3 for the 2nd.
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
1116623
y=x^2 - 4x - 5\n\nFunction:\n\ny= 1(x-2)^2 - 9\n\nNow it's up to you to find out the rest. I not going to tell you everything.
665576
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
905417
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
26326
A procedure for finding the quadratic factors for the complex conjugate roots  of a polynomial P(x) with real coefficients.\n\n[x-(a+ib)][x-(a-ib)]=\nx^2+2ax+(a^2+b^2)=\nx^2+Bx+C.
817524
For a quadratic equation, the maximum or minimum will be the vertex. Since the first term of the quadratic is positive, the parabola goes up, so you're looking for the minimum.\n\nThere are 2 methods:\nCalculus (using derivatives):\nf(x) = x2 - 2x - 4\nf'(x) = 2x - 2\nthe maximum/minimum will be where f'(x) = 0.\n2x - 2 = 0\nx = 1\ny = 1^2 - 2*1 - 4 = -5\n\nAlgebra:\nThe vertex can be found using the formula:\n(-b/2a , -(b^2 - 4ac)/4a)\n(-(-2)/2, -(4 - (4*1*-4))/4)\n(1,-5)\n\nCheck: x = 1, y = 1^2 - 2*1 - 4 = -5
745544
the quadratic formula is a useful tool in solving an equation involving a variable and its square without breaking it down into smaller pieces. It can only be used when it is in the\nax^2+bx+c=0  format. (a, b, and c can be negative.)\n\nIn this problem, a=2, b=-2, and c=-1. The equation for the quadratic is:\n\nx= (-b+/- sq.rt (b^2-4ac))/2a\n\nSolving for the equation given, you get:\n\nx= (2 +/- sq.rt.(4-4*2*-1))/4\n   = (2+/- sq.rt 12)/4\n   = 1.366, 0.366
872966
1.) 3 &#92;/ x+2 - 4 = 3\nif you mean\n3 sqrt( x+2) -4 =3,\nthen\n3 sqrt(x+2) =7\nsqrt(x+2) = 7/3\nx+2 = 49/9\nx=49/9 -2\n\n2.) &#92;/ x+3 - &#92;/2x-8 = 1\nsqrt(x+3) -sqrt(2x-8) =1\nsqrt(x+3) = sqrt(2x-8) +1\nx+3 = 2x-8 +2 sqrt(2x-8) +1\n-x + 10 = 2 sqrt(2x-8)\nx^2 -20x + 100 = 4 ( 2x-8)\nx^2 - 28x  +132 =0\n'
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
838574
Part 2 is easy.  The function y = 1/(1+x) has an infinite discontinuity at x = -1.  Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote.  For positive x, y(x) > 0, and for large negative x, y(x) < 0.  It's easy to draw the graph.\n\nIn this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2.  We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2).  Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:\n\n(1)  For x = -3:  9a - 3b + c = -1/2\n(2)  For x = -2:  4a - 2b + c = -1\n(3)  For x = 1:  a + b + c = 1/2\n\nThe solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2.  (Details of the solution omitted.)\n\nThe quadratic is y = 1/4 x^2 + 3/4 x - 1/2.  This function passes through the three points we identified, and\n\n1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem.  (Actually, there's equality at three points.)\n\nThat inequality can also be written as\n\n4/(1+x) < x^2 + 3x - 2\n\nI'll answer the other part of the question as best I can in the morning.  Right now, it's time to sleep.
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
860831
I am guessing your function to be one of the following:\n(1) f(x) = ½x - 3 or (2) f(x) = -½x -3\nIn function (1):\nf(0) = ½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = ½(2) - 3 = 1 - 3 = -2 ... or... (2, -2)\n\nIn function (2):\nf(0) = -½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = -½(2) - 3 = (-1) - 3 = -4 ... or ... (2, -4)