If  X * Y = K= a constant yield an hyperbola ,Does Y*X=K*Z  Yield the same graph.?
80323
No. xy = k is an hyperbola in 2D. xy = kz is an hyperboloid on 3D.
872619
general equation of hyperbola \n\nx^2/ a^2 or b^2 - y^2/ b^2 or a^2 = 1\n\nleaving a vertex of 1 = a it means a is in y and and the parabolas are opening up or down.\nfoci of (-1,1) means c = 1 and the center is not (0,0) it's in (-1,0)\n\nand e= a/directrix\n3/2= 1/d\n3d=2; d=2/3\nc^2=a^2+b^2 1=1+b^2 b=0\n(x+1)^2 over 0-y^2=1\ninfinity therefore \nno solution lol! sorry i know it this way
1134068
No, \nfor f(x)=(x+5)/(x-9), you substitute 1 for x, then solve for f(x) to obtain the y-value at x=1. So:\n\n                                 f(1)=((1)+5)/((1)-9)= -6/8\n\n                               which shows that at x=1, y= -6/8 = f(1)\n\nThe point (1, -5/9) is not on the graph.
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
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
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.
528695
x to the n + y to the n = z to the n where n is an integer? ... damn it's been so long I can't remember for sure. Some Brit solved the proof already.
932980
y = arcsin(cosx) is the same as\n\nsiny = cosx\n\nWhich you can show to be equivalent to x+y = pi/2 in the interval 0<=x<=pi (Hint: start with the equality sin(x+y) = sinxcosy + sinycosx).\n\nYou can also show it to be y - x = pi/2 in the interval -pi<=x<=0.\n\nThis being cyclical, you'll end up with a nice zigzag graph.  If you allow multiple values for the arcsin function, you can even litter the whole plane with lines.
421674
y = (x mod 5) + 1.\n\nThough that does seem a bit like cheating :)
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.
1126609
Actually, no.\n\nIf dy/dx = 2, then I'd say yes, in moving from one coordinate pair to another, the y-value rises twice as fast as the x-value.  (That would be the case for a line with a slope of 2.)\n\nBut when dy/dx = 2x, as it does with a curve like y = x^2 + C, the ratio (of movement in the y-direction to movement in the x-direction) isn't constant -- it varies depending on where you are on the curve.  At the point (1, 2), for instance, that "slope," a.k.a. the slope of the tangent line at that point, is 2 -- the ratio of y-movement to x-movement is 2:1.  But at the point (3, 9), the slope of the tangent line is 6.\n\nHope that helps clear it up.  The important thing is that the slope of the tangent line is changing depending on where you are on the curve.
1095247
For a fixed a and n, and allowing b and c to vary, the curve would be a hyperbolar centered at the origin. You will only be taking the rational points on that curve, though.
128537
It's arbitrary.\n\nWe are used to "x" and "y" (and perhaps "z") being used as the variables, but we also need to have a representation of constants. Using a specific number would make it seem as though only that number works in the equation.
782645
1/x=y+3\nit resembles the equation of the asymptotes of hyperbola
866301
because y- x > 0\n\n|y-x | = y- x (positive value)\n\nx-y < 0\n\nso | x-y| = y-x (as absolute value positive)
765213
yes....infact...its easy tomake linear graphs be parallel\n\nlinear curves have the equation:\ny=mx+b\n\nfor a line to be parallel they need the same slope (m)\n\nso you can change your y-intercept (b) and those two lines will be parallel\n\nfor example y=2x+4 and y=2x-1000 are parallel because they have the same slope (m)\n\nhope this helps\nmatttlocke
846505
Well...\nIf Y/Y'=k/y then Y'/Y=y/k\nso (lnY)'=y/k\nIntegrating\nlnY=ln A+ (y^2)/2k      A- constant\nExponentiating\nY=A*exp( (y^2)/2k )\nAm I right?