Find the maximum or minimum of the following quadratic function: y = x2 - 2x - 4?
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
766935
The easy way is to note that both numbers have to be equal so 56/2=28\n\n\nthe hard calculus way is to maximize xy given that y=56-x\nx+y=56 implies\nxy=x(56-x)=56x-x^2= to find maximum of 56x-x^2\ndifferentiating with respect to x gives 56-2x. The critical point is when  56-2x=0 so x=28.y=28
1049820
The quadratic formula:\nx= -b(+/-) the square root of (b squared - 4ac), all of this over 2a
456859
y = mx + b\n0 = 2m + 4\n2m = -4\nm = (-4/2)\nm = -2\n\ny = -2x + 4\n\nStandard form of this would be \n\n2x + y = 4
1072562
I'm reading the equation as \n\n4x^2 + 8x -1 = 0\n\nUsing quadratic formula x = (-8 +/- sqrt(64+16))/8 \n\nso x = -18/8 or 2/8 = -9/4 or 1/4\n\nThose are the roots.\n\nSubstitute each of those values as y into the second formula.
1008918
I'm not sure what the two pairs of max/min values in your equation are supposed to represent.  The simple calculus method for finding max/min values works like this:\n\nf'(x)=8x-4=0   take derivative with respect to x, and set = to 0\n\n8x=4\nx=.5      solve\n\nf''(x)=8      take second derivative, since it is is positive, the value is a minimum\n\nplug into equation  f(.5)=4(.5)^2 -4(.5) = -1\n\nkeep at it\nmath is good stuff
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
538847
For any quadratic of the form ax^2 + bx + c = 0, the roots are\n\nx = (-b +/- sqrt(b^2-4ac)) / 2a\n\nSqrt here is shorthand for "the square root of".
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
91768
Quadratic Functions have the form y = ax^2 + bx + c\n\nThe vertex will always have x = -b/(2a). You can calculate this number and plug it into your equation to solve for y. If a is positive, the graph will go up in both directions from the vertex. if a is negative it will go down. You can pick some x value different from the vertex to help you plot other points to allow a symmetric parabola graph.\n\nGood Luck!
807339
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1\nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n.
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
730173
x = (-b +/- sqrt(b^2 - 4ac))/2a\n\nwhere a= 1, b= -6 , c=8\n\nx1 = (6 + sqrt(36 - 32))/2 = 4\nx2 = (6 - sqrt4)/2 = 2
1146893
x^2+mx+1/x^2+x+1<3\n\ncase 1 : for x>0\n=> (x+1/x)^2 +(m+1)x <4\n=> (m+1)x < 4- (x+1/x)^2\n\nNow, we find the maximum value of the RHS:\nthe minimum value of (x+1/x)^2 is 2 when x=1 and for all other values of x it is >2 \nThis makes max(RHS)=0 or m+1 <0 \nor m<-1\n\ncase 2: x<0\nthe same argument as above(but x=-1)  leads us to m+1 >0 \nor m>-1\n\ncase 3: x=0\nThe given expression is undefined \n\nI guess its a home work problem, which indeed makes you lazy!
322681
For the first, if one side is x, then another side would be 9-x (so that they would add up to 18).  The area is then\n\nA = x(9-x)\n= 9x - x^2\n\nTo find the maximum area, take the first derivative and set it to 0:\n\nA' = 9 - 2x\n0 = 9 - 2x\n2x = 9\nx = 9/2\n\nSince the parabola opens down, the maximum area happens when x = 9/2.\n\nSo if x = 9/2, the other side is 9 - 9/2, or 9/2 - it turns out to be a perfect square!  And the area is then (9/2)^2, or 81/4.\n\nFor the second, you already have an area and you need to find the perimeter.  So, lw = 36, and we're trying to find the smallest 2l + 2w.  Well, l = 36/w, so the perimeter can be expressed as\n\nP = 2l + 2w\n= 2(36/w) + 2w\n= 72/w + 2w\n\nWe can take a derivative and set it to 0 to find the minimum value here:\n\nP' = -72/w^2 + 2\n0 = -72/w^2 + 2\n72/w^2 = 2\nw^2 = 36\nw = 6\n\nSo if w = 6, then l = 6 as well, and the perimeter would then be 24.
1120747
THE ANSWER IS FALSE. f(2) is a maximum, not a minimum. To prove this, you have to find to find f(1), f(2), and f(3). The answers are (1, 8), (1, 9), and (1,8) respectively.\n\nCALCULUS JUNK:\n\nf(x) = -(x^2) + 4x + 5\n\nTake the derivative:\n\nf'(x) = -2x + 4 = -2(x - 2)\n\nIf f'(x) = 0, the slope of f(x) is 0.\n\n0 = -2(x - 2)\n\nx = 2\n\nf(2) = 9\n\nAt (2, 9), there is a minimum OR maximum on the f(x) line. A quick graph shows that (2, 9) is a maximum.\n\n(The graph is an inverse parabola.)\n\nSo the minimum value is negative infinity.\n\nHope that helps.