what is the formula for a reflection on a graph?
405955
If you are reflecting across the y axis, x becomes -x.\n\nIf you are reflecting across the x axis, y becomes -y.\n\nIf you are reflecting about the origin, both x and y change sign.\n\nIs that what you are looking for?
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
505113
Reflection off of smooth surfaces such as mirrors or a calm body of water leads to a type of reflection known as specular reflection. Reflection off of rough surfaces such as clothing, paper, and the asphalt roadway leads to a type of reflection known as diffuse reflection.\n\nGood luck doing your homework!
741917
This one is tough.  I've looked at it three different ways and still don't get it.\n\nFirst try.  If P(x1, y1) and Q(x2, y2) are given, and y^2 = 4ax, then the derivative is 2y dy = 4a dx  ==>  m = dy/dx = 2a/y.\n\nSo (y - y1) / (x - x1) = m1 = 2a/y1.  Solve for y to get\n\ny = (2a/y1) (x - x1) + y1  (equation of a tangent line)\n\nRepeat for (x2, y2), then set the two y-values equal to each other, and solve the resulting equation for x at the point where the two lines intersect.  That result is\n\nx = [y2 (y1y2 + 2ax1) - y1 (y1y2 + 2ax2)] / [2a(y2 - y1)]\n\nso it's possible to find the point of intersection T, but the algebra is very messy.  That's where I stopped.\n\nSecond try.  Parallel rays of sunlight striking a parabolic mirror will get reflected off the tangent and redirected to the focus.  Snell's law says the angle of incidence equals the angle of reflection.  That gave me a lot of angle relationships, but I didn't make much progress.\n\nThird try.  I got it to work for one special case.  Let P(a,2a) (the top end of the latus rectum) and Q(4a, -4a) be given.  Then the slope of the tangent is given by m = 2a/y, and\n\nm1 = 2a/y1 = 1 and m2 = 2a/y2 = -1/2\n\nFrom this we find that the two tangent lines intersect at T(-2a,-a).\n\nTo get the angles POT and QOT (where O is the focus at (a,0)), I got three vectors OP = <2aj>; OQ = <3ai - 4aj>; and OT = <-3ai - aj>.  The magnitudes are OP = 2a, OQ = 5a, OT = a sqrt(10).  Then using the inner (dot) product:\n\nOP dot OT = <2aj> dot <-3ai - aj> = -2a^2 = 2a^2 sqrt(10) cos POT, so cos POT = -1/sqrt(10) and angle POT = -108.435 degrees.\n\nOQ dot OT = <3ai - 4aj> dot <-3ai - aj> = -5a^2 = 5a^2 sqrt(10) cos (QOT), so cos QOT = -1/sqrt(10) and angle QOT = -108.435 degrees.\n\nIn this special case, the angles subtended by PT and QT are equal.\n\nAnd that's better than nothing.
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!
826055
First pair both x values are the same so it is a vertical line through -5 on the x-axis. Which would mean it has no slope (undefined)\n\nSecond pair has identical y-values so it is a horizontal line (slope=0) through 3 on the y-axis\n\nFor the third pair us the formula \ny=mx+c which describes any straight line (m=slope, c= a constant distinguishing the line from others with the same slope in different locations on a graph)\n3=m(-8)+c      (a)\n2=m(-6)+c      (b)\nSimultaneous equations!\nc=3+8m           (a)\n2=-6m+3+8m   (sub (a) into (b))\n2m=-1\nm=-1/2  \nSo the slope = -1/2\n\nThis formula can be used for any problem like this using the same method.
850129
Firstly, determine the slope or gradient of the lines;\n\nLine 1:\n\nm1 = (y2-y1)/(x2-x1)\n     = (3 - -7)/(1 - -4)\n     = 10/5\n     = 2\n\nLine 2:\n\nm2 = (y2-y1)/(x2-x1)\n    = (10 - 6)/(4 - 2)\n    = 4/2\n    = 2\n\nBoth gradients are the same, therefore the lines are parallel. if the product of the gradients (m1m2) was -1 then they would have been perpendicular.\n\nWe can take this further by checking to see if they are the same line.\n\nThe equation y - y1 = m(x - x1) will be the same for both lines if this is the case.\n\nLine 1:\n\ny - 3 = 2(x - 2)\ny = 2x + 1\n\nLine 2:\n\ny - 10 = 2(x - 4)\ny = 2x + 2\n\nThese are not equal, however we now know that line 2 sits 1 unit higher on a graph than line 1, and both lines are parallel to each other.
196942
wow, that's pretty amazing.  i was convinced that was wrong until i tried drawing the "counterexample" i had in my head.  other than reading through that proof (props to the guy who posted that link, btw) i guess the best way to realize this is true is to show that for any five closed shapes in two dimensions, there will always be at least two shapes that do not touch one another.  because of this, you can color those two shapes the same color, so you will have a maximum of four colors.\n\nthis problem is reducible to a graph with five nodes, where each node represents a shape and each edge represents a shared boundary.  (showing that this is reducible to this is probably the difficult part of the proof, which is why i'm glossing over it).  you cannot connect each node to every other node without there being at least one set of intersecting edges.  if edges intersect, it indicates that the shapes (or countries) represented by the two nodes connected by one of the intersecting edges do not share a border.\n\nhere's my suggested proof (assuming we accept the reduction to this graph problem):\nstart with a graph G with three nodes, {A, B, C}.  There are also edges {AB, BC, AC}.  So, G is a closed graph with three nodes and three edges where each node is connected to each other node in the graph (i.e. G is a triangle).  If our reduction of the coloring problem is correct, then there exists a map reducible to this graph that requires three colors to be colored in the described way.\n\nNext, construct a graph H by adding a node D to graph G.  There are two possible ways to add D: 1) D can be on the interior of closed graph G; or 2) D can be outside the interior of G.\n\nIn case 1, let the new graph with nodes A,B,C,D be called H.  We need to show that we can make H such that D is connected to A, B, and C without any intersecting edges.  This is simple enough; straight edges to A, B, and C from D do not intersect each other or the existing edges of the graph G.  So, H demonstrates that a map can require four colors.  H now has four separate enclosed interior regions: triangles ABD, ACD, BCD, and ABC.\n\nFrom H, construct a graph I by adding another node E.  Repeating the above procedure, E can be placed in four places: a) in the interior of ABD; b) in the interior of ACD; c) in the interior of BCD; or d) on the exterior of ABC.\n\nCASE a: As with constructing H, edges can be drawn from E to A, B, and D.  However, C is on the exterior of ABD, so any edge from E to C must cross either AB, BD, or AD.  So, in case a, there must be an intersection, so four colors is still an option.\nCASES b and c follow similarly.\nCASE d: E is on the exterior of ABC, but D is on the interior of ABC.  Any edge between E and D must intersect AB, BC, or AC.  So, coloring with four colors is possible in this case.  Since it is possible in all cases, it is possible for any map with five regions with a sub-map reducible to a graph like H.\n\nCase 2) Instead of placing D in the interior of G, make a graph H' by placing D on the exterior of triangle ABC.  Any quick drawing will show that D is easily connected to A, B, and C.  So there is a map with four regions that requires four colors.  To make the rest of this easy, note that H' also produces four enclosed regions: ABD, ABC, BCD, and ADC.  I won't go through the trouble to prove this, but clearly H' is isomorphic to H.  That being said, whatever is true of the general graph H is also true of the general graph H'.  Therefore, a graph I' made by adding a node to H' cannot be complete without there being at least one set of intersecting edges.\n\nSince this is true for all cases, any map reducible to a graph with five nodes is colorable with just four colors.  QED\n\nSorry this is so long.  I hope it's helpful.
635075
Graph X=-4 with dotted lines\n\nthen color everything below the line.
155791
Here is a good page describing slope.\n\nhttp://mathforum.org/cgraph/cslope/slope.html
273336
try this http://www.accessexcellence.org/RC/VL/GG/diagram.html\nAll the best !!
921732
If you look at a picture (aka a graph) of the function, it will all become clear...\n\nhttp://upload.wikimedia.org/wikipedia/en/3/3e/1-over-x.png
914601
http://mathworld.wolfram.com/PoincareConjecture.html
416187
I think he means like, math-related reflections.\n\nA reflection is like a flip of the figure.
1055471
You need to graph this function. \nIt is a line.\ny = (x^2)^ 1/2 - 16 \ny = x - 16\nx = 0, y = -16, \nx = 16, y = 0
44938
Let \nBlue = x\nYellow = y\n\nThen:\n2x + y = Green   [Given]
337864
reflection of the sky