Find the four (4) intersections of the parabolae:?
83507
If you plot the two parabolae very carefully, you will see that they do not intersect, at least in real space. However, if you allow complex roots to the resulting quartic equations, then one can of course find 4 complex roots. \n\nso y takes the values:\n\n-6.0583   +   i*0.5986\n-1.9416  +   i*0.3398\n-6.0583   -   i*0.5986\n-1.9416  -  i*0.3398\n\nand at each of those values of y, x takes the corresponding values:\n\n   3.8783 - 2.4642i\n   4.1215 + 1.3989i\n   3.8783 + 2.4642i\n   4.1215 - 1.3989i
949810
Use the distance formula:\n\nD = √ [ (y2-y1) ² + (x2-x1) ² ]\n\n1)  D = √ [ (-2+4) ² + (-2+4) ² ] = √(4+4) = √8 = 2√2\n\n2)  D = √ [ (3+1) ² + (-2+7) ² ] = √(16+25) = √41\n\n3)  D = √ [ (4-5) ² + (6+3) ² ] = √(1+81) = √82\n\nNRT
184897
Divide this equation by 4 on both sides.\n\nYou get \n\n3y^2-x^2+18y+4x+44 = 0\n\nso, (3y^2+18y+27)-(x^2+4x+4)-12 = 0\n\nso, 3(y^2+6y+9)- (x^2+4x+4)-12 = 0\n\nor, 3(y+3)^2 - (x-2)^2 = 12\n\nThis is clearly a hyperbola.\nNow, what you have to do is take Y = y+3 and X = x-2 (note the difference between caps and lower case letters) and divide the whole equation by 12\n\nSo you get Y^2/(2)^2 - X^2/(2sqrt3)^2 = 1\n\nhence a = 2, a^2= 4, b = 2sqrt3, b^2=12, c^2=a^2+b^2=16 and c = 4.\n\nVertices = (0,2),(0,-2) in the X,Y system so it is (2,-1) and (2,-5) in the x,y system. Remember x= X+2 and y = Y-3\n\nFoci = (0,4) and  (0,-4). So, they should be (2,1) and (2,-7) in the x,y system.\n\nEquation of asymptote lines = \n\nY = (X)(1/sqrt3) and Y = -X(1/sqrt3)\n\nso they will be (y+3) = (x-2)(1/sqrt3) \n\nor (y+3) = (2-x)(1/sqrt3)\n\nThe other things can easily be calculated.
512252
Standard form of a parabola that opens left or right is\n\nx = c(y - h)^2 + k\n\nFor your equation, complete the square to get it in standard form:\n\nx = 3y^2 + 6y + 4\n= 3(y^2 + 2y) + 4\n= 3(y^2 + 2y + 1) + 4 - 3\n= 3(y + 1)^2 + 1
227387
D = 36 - (4)*(-2)*(-3)= 36- 24 = 12\nTwo real solutions.
771223
Strange parabola.\n\nBring the form to x = a(y-h)^2 + k\nThe vertex is (h,k).\nThe directrix will be x = k - m\nwhere m = 1/(4a)
906157
foci: (h, k ± c) h = 0, k = 0, c = 8  (0, 0 ± 8)\n\nasymptotes: y - k = ± (a / b)(x - h)\ny - 0 = ±(a / b)(x - 0)\ny = ±(a / b)x\na / b = 4\n\nc^2 = (4b)^2 + b^2\n64 = 16b^2 + b^2 = 17b^2\nb = √(64 / 17) = 1.94\n\na = 4b = 4(1.94) = 7.76\n\n(x^2 / 60.24) - (y^2 / 3.76) = 1
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
1057389
Look at the coefficient of the x^2 term.  If it is positive, the parabola opens up and has a minimum value; if it is negative, the parabola opens down and has a maximum.  You have 2x^2, so you will have a minimum value.\n\nNext find the vertex.  The x-coordinate of the vertex lies at -b/2a.  So -8/(2)(2) = -8/4 = -2.  Plug this value in to get the y-coordinate of the vertex.  2(-2^2) + (8)(-2) + 5 = 8 - 16 + 5 = -3.\n\nSo you have a minimum value of -3.
831482
We just started this stuff in school.  first, think of f(x) as y.  then make a table with the left column as x, and the right column as y.  the vertex is going to be -b/2a.  so in this case.  it'll be -1/4.  that's the middle number in the table and also the vertex.  now put above that number in the table on the x column as so.\nx  y\n1\n0\n-1/4\n-2.5\n-3.5\nnow fill in the the x column by plugging in the x variables.  plug in 0 for x and thats your y-intercept.  then plug in your 0 for y, and thats your x-intercept.  this is going to form a parabola so the axis of symmetry is -1/4.  the domain is going to be all real numbers because there are no contradictory numbers to fill in.  Im not sure how to do the range because im only a junior in high school but I hope I helped a little.
965380
(x - 3)² + (y + 2)² = 4²\n(C): x² + y² - 6x + 4y - 3 = 0
760976
The parabola y = (x+3)(x-5) crosses the x axis at -3 and 5.  It opens upward, so its vertex must be below the x axis.  So y < 0 for x in (-3, 5).
798759
1- yes, 2-yes, 3-no Vx=the same 24.5m/s, 4-yes, 5- no Dy=21, 6-yes Ax=0 throughout the whole parabola. 7- yes coz the only force (constant force) is gravity. Now (8) – its final Vy=26, and Vx=  still the same 24.5, thus Vtotal=sqrt(Vx*Vx+Vy*Vy)
619069
the parabola will be at the point 2,2 and shaped like a "u"
384702
The first 3 are all straight lines, so you just need to get 2 points in each.\n1. y = |x| + 3 for x = 1 then y = 4, for x = 2 y = 5.\n\n2. y = 3/2x - 2 for x = 2 then y = 1, for x = 4 y = 4\n\n3. y = x - 5 for x = 5 y = 0 for x = 6 y = 1\n\nFor #4. This is a curve shaped like and upside down U that passes through 0,0, for all values of x, y will always be negaive.\n\nThe best way is to substitute values of x and get values of y and sketch.\n\nHope I was some help
41570
A parabola with its base at 0,1.  Go to http://www.coolmath.com/graphit/ Enter x^2+1 and see it for yourself!  :)
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)