Which graphs have a vertex of (0,0)? Explain how you know.?
366652
The first two answers are completely wrong. The fact that (0, 0) satisfies the equation only means (0, 0) is a point on the parabola, not that it is also the vertex of the parabola. For example, plugging (0, 0) into y = (x - 1)^2 - 1 yields a true equation because the point is on the graph, but the vertex of the parabola is the point (1, -1)!\n\nRemember that if you have a function y = f(x), then y = f(x - k) for some constant k is just the graph of f(x) shifted k units to the right (see what happens to f(k) and f(0) when you subtract k from the argument).\nSimilarly y = f(x) + j will be the graph of f(x) shifted upwards by j units.\nNow, a parabola with vertex at the origin has function f(x) = hx^2 for some scaling constant h. The forms we are looking for are then y = h(x - k)^2 + j where the vertex will have been moved to (k, j). Thus, we just need to adjust our equations to the form above and read (k, j) off of them.\nFor example, your first equation is y = 3x^2 - 4. Adjusted to the form above, we get y = 3(x - 0)^2 + (-4). Thus (k, j) = (0, -4), so this graph's vertex is not at the origin.
323669
Well, this isn't true, actually.  The graph on six vertices consisting of two connected components, each a triangle, is 2-regular but not hamiltonian.  This isn't even true for connected 4-regular graphs.  For example, take two copies of the Petersen Graph (go to Mathworld if you don't know what this is) and add 10 edges joining vertices in one copy with their "twin" vertex in the other copy.  This graph is 4-regular and not hamiltonian.
779402
Well, its very easy to do this. When you have 2 different graphs, you graph then on your graphing calculator, then wherever the two graphs intersect, is the solution to the system. So if there are more than 1 point, means there are more than one solutions. IF you have more than 2 graphs, then you have to find the intersection  point where all of them meet and that would be the solution for the graphs.
497181
You only need two points to determine a line.\nYou should use three just to make sure you don't make a mistake.\nPick two values of x and find the corresponding values for y.\nThis one is pretty simple.\nLet x = 0, then y = 0 (since y = x)\nLet x = 1, then y = 1\nLet x = -3, then y = -3.\nPlot these points on the graph and connect them with a straight line.
788027
The graph is a V.  From the 0,0 draw a line with slope = 1 to the right of the y axis and draw a line with slope = -1 to the right of the y axis.  For y<|x|.  All you have to do is shade in the area below the V.  If y>|X|, all you have to do is shade in the area above the V.  Less than, shade below; greater than, shade above.  That should answer all the questions you've submitted.\n\nAnd no, absolute values do not form parabolas.  Some do, not all.  Like y = sqrt[|x-1|].  This forms 2 graphs each going away from each other.\nNice going NerdyKid.  You're not a nerd, you're a dork.
916256
first express the equation in the form\ny=a(x-h)^2+k\nif the a is positive itwill open upwards\nif a is negativeit will open downwards\nfor finding the y intercept put x=0\nin this form y=ax^2+bx+c y intercept will bw (0,c)\nthe vertex will be given by(h,k)
597648
y= -4x^2+8x-1\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
717431
On a graph, these are the points where the line you are drawing crosses the "x" and "y" axis.  It's hard to give an example without drawing a graph, but you can see some here:\n\nhttp://www.tpub.com/math1/13a.htm
206626
Do you mean graph? Look in your math book. You take a pencil and you graph ordinarly basically.
919538
it is narrow because a is more than 1.\nThe vertex=1,-1 you take the opposite of h which=x and y=-1\ndo a t table to find the max and min \nput the vertex in and put points for x and solve for y for each one using the equation
133564
The coordinates of the vertex of a parabola \ny = ax^2 + bx + c is\n[-b/2a , (4ac - b^2)/4a]\nJust substitute and you will get\nV(2 , -7)
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).
844444
Remember that the vertex of the parabola for y = x² is at (0,0).  You can confirm this by putting x = 0 in and seeing that you get y = 0.\n\nIf you instead replaced x with (x - 3):\ny = (x - 3)²\n\nNotice how it would take a value of x = 3, to get y = 0.  In other words, the whole graph, including the vertex would move 3 units to the right.  The new vertex would be at (0, 3).\n\nSimilarly, if you replaced y with (y - 2):\ny - 2 = x²\n\nThis is the same as:\ny = x² + 2\n\nSo every value would move up 2.  The graph would move two units up and the vertex would similarly move to (0, 2).\n\nCase 1:\nParabola moves 3 units to the right -- vertex now at (3,0)\n\nCase 2:\nParabola moves 2 units up -- vertex now at (0, 2)\n\nExtra credit question for you:\nWhat would happen if you did both?\n(y - 2) = (x - 3)²
794182
What nicknamr said, and also, the vertex would be located at the x-value found for the axis of symmetry, and the y-value for that x-value.  So, -b/(2a) = -12/(2 x -1) = 6.  Then, y= -(6)^2 + 12(6) -32 = -36 + 72 -32 = 4.  The vertex comes out to be (6,4).  (I would have answered all the other stuff too, but he already did it, and I found it rather redundant to repeat it all, so I just added on to it =)  )
632855
4. x^2 - 3x <= 0\n\nif you factor:\n\nx(x - 3) = 0\n\nwhich means either\nx = 0 or x-3=0\nx=0 or x=3 these are the endpoints represented on the graph. if you want the area enclosed (underneath) the graph, it has to be <= 0
184980
4x-2y-6=0\n+6      +6\n4x-2y=6\n\ny-2x=-5\n*2   *2\n -4x+2y=-10\n+ 4x-2y=6 (adding the previous equation)\n  0+0=-4\n0=-4\nNo solution\n\nTo graph:\ny-2x=-5\n+2x  +2x\ny=2x-5\nGraph that\n\n4x-2y-6=0\n+6      +6\n4x-2y=6\n-4x  -4x\n-2y=-4x+6\n/-2   /-2\ny=2x-3\nAnd graph that. As you can see, the lines do not intersect, confirming the answer of no solution.
890433
go here- http://people.hofstra.edu/staff/steven_r_costenoble/Graf/Graf.html\n\nand enter     2x^2   in the big blank field and a graph will pop up!