How would I write this in a quadratic function for this parabola?
958976
Here is how:  The axis of symmetry occurs halfway between the x-intercepts.  That is x = 1.  So the vertex is (1,6).\n\nThe vertex form of a parabola's equation is y = a(x-h)^2 + k\nwhere (h,k) is the vertex.\n\nSo we know it's y = a(x-1)^2 + 6\n\nFinally to find a, plug in the known coordinates of a point other than the vertex, say (4,0) since those are the coordinates of an x-intercept.\n\n0 = a(4-1)^2 + 6\n\n0 = 9a + 6\n\na = -6/9 or -2/3\n\nSo the equation is y = -2/3 (x-1)^2 + 6\n\nIf you need a different form, multiply it out but this should suffice.  Hope that helps explain what to do.
296452
for a parabola, y = ax^2 + bx + c, so plug in the x,y coordinates of the points:\n\nP:  3 = a(7^2) + b(7) + c\nQ:  7 = a(9^2) + b(9) + c\nR:  1 = a((-2)^2) + b(-2) + c\n\nyou get:\n49a + 7b + c = 3\n81a + 9b + c = 7\n4a - 2b + c = 1\n\nsubtract equation 2 from eq 1 and eq 3 from eq 1 to get\n32a + 2b = 4\n45a + 9b = 2\n\nnow it gets ugly.  b has smallest coefficients, so divide eq 1 by 2 and eq 2 by 9 to get\n16a + b = 2\n5a + b = 2/9\n\nsubtract to get 11a = 16/9, then divide by 11:\na = 16/99.  plug this into the easiest equation, 5a + b... \n\n5(16/99) + b = 2/9\n80/99 + b = 2/9 = 22/99\nb = -58/99\n\nthen plug both a and b into an earlier equation:\n4a - 2b + c = 1\n4(16/99) - 2(-58/99) + c = 1\n64/99 + 116/99 + c = 1\n180/99 + c = 1\nc = 99/99 - 180/99 = 81/99 = 9/11\n\nso the parabola is y = 16/99 x^2 - 58/99 x + 9/11\n\nwasn't that fun?
737893
Take your quadratic equation.\n\nThe graph cuts the x axis when y = 0\n\nSo simply put your equation = 0, and then solve it either by using factorising or the quadratic formula:\n\n(-b plus or minus sqrt(b^2 +4ac)/2a)\n\nif I remember right, (could be minus 4ac)
202779
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!
1049820
The quadratic formula:\nx= -b(+/-) the square root of (b squared - 4ac), all of this over 2a
838654
I would factor each quadratic and then see if I could somehow cancel some stuff out.\n\nx² - x - 6 = (x-3)(x+2)\ny² + 3y - 10 = (y+5)(y-2)\ny² - 7y + 10 = (y+5)(y+2)\n3x² + x - 10 = (3x-5)(x+2)\nx² - 2x - 3 = (x-3)(x+1)\n9x² - 25 = (3x+5)(3x-5)\n\nRewrite your problem with these factors and you will see that you can cancel out a lot of these terms!
848564
Substitute 2x-1 for y in the middle part, you get:\nx^2 + x -6(2x-1) = 2x-1\nThis you can reduce to a quadratic equation for x.
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
780298
first u distribute the - to the stuff in the parenthesis so you have\n7x^2-3+x^2\nthen combine like terms\n8x^2-3
254355
Things you need to know\n\nFormula for a parabola (with vertical symmetry): (x-h)^2 = 4a(y-k) where "h" and "k" are the x and y corrdinates of the vertex, and "a" is the distance from the vertex to the focus.\n\nCoordinates of the vertex: (h,k) - these tell you how far to shift your parabola from the origin.\nCoordinates of the focus: (h, k+a) - if a is the distance from vertex to focus, then k+a should give you the point of the focus\nFormula of the directrix: y = k-a -- the vertex lies halfway between the focus and directrix, so subtracting the distance from the vertex to the focus will give you the directrix.\n\nPlease ask your teacher if you do not understand the reasoning behind any of these formulas.\n\n~~~~~~~~~~~~~~~~~~~~~\nNow, from what you are given, a vertex at (0,0) and a focus at (0,4) you can see that since the y coordinate is the one that changes, the focus lies above the vertex (not left or right) so it must be a vertically symmetrical parabola.  You can also tell that it opens upward, since the focus lies inside the parabola and it is above the vertex.  This means that the directrix is a flat line below the parabola.\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nOn to solving your problem:\n\nVertex (h,k) = (0,0)\nFocus (h,k+a) = (0,4) so k+a = 4 -- from the vertex you know that k=0, so 0+a = 4, or a = 4\n\nRemember the formula for the directrix was y = k-a.  Plug in what you know and you get y = 0-4, or y = -4\n\nIf this still doesn't help, let me know and I'll try again :)  good luck!
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
870960
Do you mean that the roots of x^2+pX+r are α  and β and you need to find quadratics in terms of p and q?\nIf so you can use the following:\nIn any quadratic ax^2 + bx + c = 0, the sum of the roots = -b/a and the product of the roots = c/a.\nUsing this,  -(α + β) = p, and αβ= r.\n\nAnswers are for terms b and c in the quadratic x^2 + bx + c\ni)\nb = -(1/α + 1/β) = -(α + β)/αβ = p/r\nc = 1/αβ = 1/r\n\nii) \nb = -(1/α^2 + 1/β^2) =  -(α^2 + β^2)/(αβ)^2 = -[(α + β)^2 - 2αβ])/(αβ)^2 = -(p^2 - 2r)/r^2\nc = 1/(αβ)^2 = 1/r^2\n\niii) \nb = (α^2β + αβ^2) = αβ(α + β) = -pr\nc = -α^2β * -αβ^2 = (αβ)^3 = r^3
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0
418438
Um.  I don't really understand the first answer above.\n\ny+4 = +- sqrt(4(x-3))\ny = -4 +- sqrt(4(x-3))\n\nI guess this is a sideways looking parabola that points to the right.  It's shifted down 4, to the right 3, and has a scaling factor of 1/4.
65888
a parabola facing up!\n\nwith center (0,1)
259902
When you solve the quadratic eqn, you are solving for x. You should get 2 values for x, these are where your parabola crosses the x axis. \nYou need to know it for a number of reasons, most of which become more obvious as you move into higher levels of math and science.\nThe path followed by a parabola could represent a number of things...for example, it could be the height of a thrown object object (y axis) vs. the time that the object has been in the air (x axis). Solving the quadratic to this kind of question will tell you the 2 x values, which would represent at what time the object was thrown and what time it hit the ground. The vertex of the parabola (-b/2a) would be the time at which the object had reached it's max height.
949364
since the second derivative function is a constant, it follows that the function is quadratic of the form ax^2 + bx + c = f(x)\nf(1) = 1 so a+b+c=1\nf'(x) = 2ax+b => f'(1) = 2a+b = 3 => b=3-2a\nf''(x) = 2a = 2 so a=1\n=> b=3-2(1) = 1\n=> (1) + (1) + c = 1 => c=-1\n\nf(x) = x^2 + x - 1\n\ni assume you made a mistake -- you said "show that f''(x)=x^2 + x - 1"\ni took that to mean f(x) not f''(x), for you just stated that f''(x)=2 for all x on the interval