Quadratic Form to Vertex Form (please help!)?
830875
You have an equation in the form of y = ax^2 + bx + c, and you want the form (y - k) = a(x - h)^2\nManipulating all those decimals can be very confusing, so let's try manipulating letters, instead.\na = -4.97\nb + 10.91\nc = -5.16\ny = a(x^2 + (b/a)x) + c\ny = a(x^2 + (b/a)x + (b/2a)^2x^2 - (b/2a)^2) + c\ny = a(x^2 + (b/a)x + (b/2a)^2x^2) - a(b/2a)^2 + c\ny + a(b/2a)^2 - c = a(2 + (b/2a))^2\ny + (b^2/4a) - c = a(x + (b/2a))^2\n(y + (b^2 - 4ac)/4a) = a(x + (b/2a))^2\nNow, substituting the numbers back in:\n(y +((10.91)^2  -4(-4.97)(-5.16))/(4(-4.97) = -4.97(x + (10.91/(2(-4.97))^2\n(y + (119.0281 - 102.5808)/-19.88) = -4.97(x + (10.91/-9.94))^2\n(y -16.4473/19.88) = -4.97(x - (10.91/9.94))^2\n(y - 0.82733) ≈ -4.97(x - 1.0976)^2
1178042
To find the x-coordinate of the vertex, use -b/2a.\n\n-6/(2)(1) = -3\n\nTo find the y-coordinate, plug this back in.\n\n(-3)^2 + (6)(-3) -2 = 9 - 18 - 2 = -11\n\nSo the vertex is at (-3, -11)
994152
First, I will say that all the gory details for the following are listed at wikipedia.org.  I'm not going to repeat them, but try to explain some of the missing steps.\n\nGiven a general cubic equation x^3+ax^2+bx+c=0, we can reduce the equation to the form \n\ny^3+py+q=0, \n\nwhere y=x+(a/3) and where p and q are somewhat complicated expressions in a,b,c (not horrible, but not worth writing out here.) This is so because, with x=y-(a/3),\n\n(y-(a/3))^3 + a*(y-(a/3))^2 + b(y-(a/3) + c\n=y^3 - a*y2 + a* y - (a/3)^3 + a* y^2 + a bunch of other stuff\n\nand so the terms in y^2 will cancel.  Work by some Italian mathematicians in the 16th century led them to believe that this was a good form into which to put the equation, because they could then solve it.\n\nNow, try substituting the expression v-u into the reduced equation.  Why?  Possibly some trial and error led the original discoverers to this notion.  What does happen is that when you make this substitution, you get the polynomial\n\nv^3 - 3v^2u + 3vu^2 - u^3 + pv - pu + q = 0\n\nRegroup a bit to get \n\nv^3 - u^3 + (v-u)*(p-3uv) + q=0.\n\nThen we could select v^3 - u^3 + q=0 and p-3uv=0.  This is a set of simultaneous equations which we can, in principle, solve for a set of values (u,v).  One way to do so is by substitution; write u=p/(3v) and substitute into the other equation.  This yields a quadratic equation in v^3 which is easy to solve, and which in turn can be solved for v.  This gives you the value of v-u, which gives you a solution to the reduced equation.  Using the relation y=x-(a/3) gives you a solution to the original equation.  I'm not writing out all the gory details because you can get all of those from a site like wikipedia or mathworld.wolfram.com.\n\nHaving said all that, it's unlikely you'd ever need to use this for anything.  One useful fact is that ANY cubic equation with real coefficients has at least one real solution (thanks to the intermediate value theorem) so knowing this root (or a good approximation) will allow you to reduce your problem to a quadratic equation, which is easy to solve.\n\nAnd yes, I'll include a link to a site which does list the gory details.
99265
I asked that same question to my teacher last time. She told me that I have to Seperated the shape into different common shapes. (for example like lets say I wanted to seperate a "square" into some common shapes I could turn it into two seperate triangles or four different triangles. Well Just try to use the formulas you know for the areas of the different disected figures you get.
148359
No true factored form.\n\na=8 b=6 c=8\nx= (-b +- squareroot[b^2-4ac])/2a\nx= (-6 +- squareroot[36-256])/16\nx= (-6 +- squareroot[-220])/16\nx= (-6 +- 2i{squareroot[55]})/16\nx= (-3 +- i{squareroot[55]})/8
516945
duff beer is incorrect.\nx = -b/2a gives ONLY THE AXIS OF SYMMETRY, not the vertex.\n\nIn order to find the vertex of a parabola (graph of a quadratic) in the form\nax^2 + bx + c = y\n\nThe x coordinate of the parabola is \nx=-b/2a\nJust plug in b and a.\n\nThen, substitute your answer for x back into the original quadratic and solve for y. You will then have the y coordinate of the vertex and thus the vertex at (x,y)\n\nAn alternate method is to complete the square on it, but that takes longer and you can research it if you want to learn more.\n\nBy the way, the above holds true only for an up or down parabola. If you have a sideways parabola, then the equation is\nay^2 + by + c = x\n\nThe y coordinate of the axis of symmetry is \ny=-b/2a\nPlug back in and solve for x and you have the (x,y) vertex.
802559
You are given the quadratic function\ny = x² - 8x + 22\n\nTo find the vertex, change the given equation into the form\ny = a(x - h)² + k\n\nWhere \na = is a constant\n(h,k) is the vertex\n\nThus,\ny = x² - 8x + 22\n\nWe complete the square, by squaring the half of the coefficient of x (which is -8, the half of -8 is -4, and the square of -4 is 16.)  Thus,\ny = x² - 8x + 16 + 6\n\nNow, we can factor\ny = (x - 4)² + 6\n\nTherefore, the vertex is\n(4,6)\n\n^_^
72405
What is "quad in form..."???
265248
the general form of a quadratic equation ax^2 + bx + c =0\n\ntherefore there is two solution\nx=[-b ± sqrt (b^2-4ac)]/(2a)\n\nx^2+2x+2=0\n\na=1 b=2 c=2\n\nx1=[-2+sqrt(2^2-4*1*2)]/2 =(-2+2i)/2 = -1+i\nx2=[-2-sqrt(2^2-4*1*2)]/2 =(-2-2i)/2 = -1-i\n\nwhere i is the square root of -1 and it is imaginary\n\nthe solution is X={-1+i , -1-i}\n\nThose solution means that this equation in the from y=x^2+2x+2 has no intersection with y=0 since no real solution for x^2+2x+2=0
997471
IF you want to try and get it in that EXACT form dicatated by Mathematica, I suggest you start with u=1+sin(x) and then solve for x in terms of u and do some resubstitution.  You'll get the arcsin function, which becomes a polynomial in terms of another dummy variable when you start the first u*dv substitution.  Notice I said "FIRST", because to get it in the form that you want, you'll have to do it about 6-10 times.  Also, you'll have to do some "right triangle simplification" throughout the problem as well.  Of course, then you get to factor it.  Have at it.  I did this very thing, stopping at the 3rd u*dv substitution because it is just ridiculous.  \nI did, however, get several of the terms.  I think if you decide to do this, you're going to need lots of paper.  I was doing it symbollically in a CAD program to keep things organized and such, and it was already 5 pages.  Oh, and not to mention that this "answer" given by Mathematica isn't even a closed form "answer".  It left it in terms of a polylogarithm, which is still an integral hidden in notation.  \n\nThe point is that the fuction (among others) is best integrated numerically because there is no useful closed form for this function.  This is what I did and it took me 2 minutes.  \n\nx^2/(1+sin(x))=x^2 - x^3 + x^4 - (5/6)x^5 + (2/3)x^6 - (61/120)x^7 + (17/45)x^8 - (277/1008)x^9 .................\n\nIntegrate termwise, which is fricking easy.  \n\n(1/3)x^3 - (1/4)x^4 + (1/5)x^5 - (5/36)x^6 + (2/21)x^7......\n\nNotice it is not undefined (like the original function) and the error term goes to 10^-7 after the first 5 terms.  I'd say that's as accurate as anyone would ever really need to be, and if not, they can always expand this function further.\n\nTell your uncle he's a bastard just for me.  He's teasing you.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
285238
you can take 9v^2 out and then it becomes a simple quadratic equation\ni can give u d ans \nit will be\n\n9v^2(v+2)(v+1)\ni hope u understood
123034
If it is y = x^2+4x+6 then proceed as follows\ny = x^2+4x+4-4+6 (add and subtract 4)\ny = (x+2)^2+2\ny-2=(x+2)^2\n\nFor vertex y-2 = 0 i.e. y = 2 and x+2 = 0 i.e. x=-2;\nSo vertex coordinates are (-2,2).
27352
the x coordinate of the axis of symmetry is given by -b/2a. here it is -(-12)/-6 = -2. plug this back into the equation and you get f(x) = 27 so your vertex is (2,27)
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!
760751
a) i)  (0,3)\nThe y-intercept is when x=0, right?  So just plug in zero for x and get:\ny=(0)^2-4(0)+3\nWe get 3 since everything else is zero.\n\nii) (3,0) and (1,0)\nOkay, we have to find when the equation is zero for the x-intercepts.  So, set the equation equal to zero (y=0)\n0=x^2-4x+3\nYou can factor this, which is finding two things that, when multiplied together, form this equation.  The two factors are\n(x-3) and (x-1)  Our factored form is\n(x-3)(x-1)=0\nOne of them MUST equal zero for the whole thing to be zero since they are multiplied, but either one can equal zero.  So we have two possible x values: 3 and 1.  Those are the x-coordinates, the y coordinates are zero.\n\niii) (2,-1)\nThe vertex is halfway in between the roots, which are 3 and 1.  You can take the average of them:  (3+1)/2=4/2=2.  The x-coordinate of the vertex is 2.  Now plug that in:\n\n(2)^2-4(2)+3 = 4-8+3=-4+3=-1\nThe y-coordinate is -1.\n\n\nb) x=2\nThis is just a vertical line with the same value as the x-coordinate of the vertex (the line of symmetry goes through the vertex).  The line is x=2
534396
quadratic equations of the form \n\nax^2+bx+c=0\n\ncan be used with the quadratic formula, see\n\nhttp://mathworld.wolfram.com/QuadraticFormula.html\n\nso just transform all your equations into the form\n\nax^2+bx+c = 0 (x can be called differently, of course).\n\nEdit: for example, let's take \n\n5t^2 + 20t - 60 = 0\n\naccording to the quadratic formula, the solutions are\n\nt_1 = (-20+sqrt(20*20-4*5*(-60))) / (2*5) = (-20+sqrt(1600))/10 = 2\n\nand\n\nt_2 = (-20-sqrt(20*20-4*5*(-60))) / (2*5) = (-20-sqrt(1600))/10 = -6\n\nNote: if the term below the root (sqrt) is negative, it is not a valid solution - sometimes the quadratic equation has only one solution, or none.