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Thread: function point

  1. #16
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    Re: function point

    How to look graphics functions

    a) y=|a-x|
    b) y=-|a-x|
    c) y=a-x
    d) y=x-a
    e) y={|a-x|} \cup{-|a-x|}
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  2. #17
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    Re: function point

    Quote Originally Posted by point1967 View Post
    How to look graphics functions

    a) y=|a-x|
    b) y=-|a-x|
    c) y=a-x
    d) y=x-a
    e) y={|a-x|} \cup{-|a-x|}
    I am not sure what you mean by this but if you mean "What do the graphs look like?" then
    a) y= |a- x|= |x- a| is a straight line with slope -1 for x< a, ending at (a, 0) then a straight line with slope 1 for x> a, starting at x= a. A "V" shape.
    b) y= -|a- x|= -|x- a| is a straight line with slope 1 for x< a, ending at (a, 0) then a straight line with slope -1 for x> a, starting at x= a. An upside down "V" shape.
    c) y= a- x. A straight line with slope -1 passing through (0, a) and (a, 0).
    d) y= x- a. A straight line with slope 1 passing through (0, -a) and (-a, 0).

    e) is meaningless- the union is defined for sets, not expressions. If you mean the union of the graphs (which are sets of points) of (a) and (b) then it is two straight lines, with slopes 1 and -1, both passing thorough (a, 0). An "X" shape. It is also the union of the graphs in (c) and (d).

    There is nothing new here. This is all pretty standard secondary school math.
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  3. #18
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    Re: function point

    a) y=|2-x|
    graph, the red surface
    https://cfxpzq.bn1302.livefilestore..../01.png?psid=1

    b) y=-|2-x|
    graph, the red surface
    https://nq6hfq.bn1302.livefilestore..../02.png?psid=1


    c) y=2-x
    graph, the red surface
    https://0nivia.bn1302.livefilestore..../03.png?psid=1

    d) y=x-2
    graph, the red surface
    https://d6pekg.bn1302.livefilestore..../04.png?psid=1

    e) y={|2-x|} \cup{-|2-x|} or y={2-x} \cup{x-2}
    graph, the red surface
    https://qhdsnq.bn1302.livefilestore..../05.png?psid=1

    which are geometric objects obtained for valuesx and y , shape a≥x≥b ( a≥y≥b ) ? , you have a graph
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  4. #19
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    Re: function point

    Each of these is wrong.

    Quote Originally Posted by point1967 View Post
    The red area (extending infinitely) is the graph of the inequality y> |2- x| The graph of y= |2- x| is the boundary of the red area.

    b) y=-|2-x|
    graph, the red surface
    https://nq6hfq.bn1302.livefilestore..../02.png?psid=1
    The red area is the graph of the inequality y< -|2- x|. The graph of y=-|2- x| is the boundary of the red area.

    The graph of y= 2- x is the slanting boundary. The red area shown above the x-axis is the graph of y> 2- x for x< 2. Below the x-axis, it is y< 2- x for x> 2.

    The graph of y= x- 2 is the slanting boundary. The red area, for y> 0 is the set of y> x- 2 for x>2. For y< 0, it is the set of y< x- 2 for x< 2.


    e) y={|2-x|} \cup{-|2-x|} or y={2-x} \cup{x-2}
    graph, the red surface
    https://qhdsnq.bn1302.livefilestore..../05.png?psid=1
    The graph of that y is the union of the two lines forming boundary. your red are is y> (2-x)(x- 2) (again, NOT "union") for y> 0 and y< (2- x)(x-2) for y< 0.

    which are geometric objects obtained for valuesx and y , shape a≥x≥b ( a≥y≥b ) ? , you have a graph
    I wonder why you keep posting these when you are paying no attention to the responses.
    Last edited by HallsofIvy; Mar 17th 2015 at 12:16 PM.
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  5. #20
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    Re: function point

    Quote Originally Posted by HallsofIvy View Post
    Each of these is wrong.

    The red area (extending infinitely) is the graph of the inequality y> |2- x| The graph of y= |2- x| is the boundary of the red area.
    it seems that you are not reading the rules I set , mapped straight line (a and x) not mapped (x) , for your test to see whether or not an innovative mathematician ,
    that geometric object in the plane meets these requirements
    a ) has six angles
    b )the sum of the angles 1080^o

    a) y_{x.}=|a_{x.}-x_{x.}|
    b) y_{x.}=-|a_{x.}-x_{x.}|, the same graph is reversed only to 180^o , and relates to a negative value y
    the scene ( x.)x-coordinates , ( y.)y-coordinates ,( xy.)plane
    Graph functions y_{x.}\rightarrow y_{y.} , mapped straight line (y_{y.},a_{x.},x_{x.})\rightarrow(a_{xy.}x_{xy.})
    2≥y≥0 ( The general form b≥y≥0 , b>0 ) rectangular isosceles triangle
    https://2bl1tq.bn1302.livefilestore..../y1.png?psid=1

    3≥y≥1 ( The general form c≥y≥b , b>0 , c>0 ) regular trapeze
    https://dc4d8a.bn1302.livefilestore..../y2.png?psid=1

    1≥x≥-1 ( The general form c≥x≥b x<a , c≥x≥b x>a ) rectangular trapeze
    https://o9amca.bn1302.livefilestore..../y3.png?psid=1

    6≥x≥-1 ( The general form c≥x≥b , b>a , c<a , |b| \neq|c|) pentagon
    https://pkxoqg.bn1302.livefilestore..../y4.png?psid=1
    more geometric objects that can be obtained ???
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  6. #21
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    Re: function point

    Operations on sets - difference, this operation returns a new geometric objects
    { 5≥ x ≥0 } \setminus{ {1≥y≥0} , hexagon
    https://nq6ifq.bn1302.livefilestore..../a1.png?psid=1
    { 3≥y≥0} \setminus{1≥x≥0} heptagon
    https://0niwia.bn1302.livefilestore..../a2.png?psid=1
    {5≥x≥-1} \setminus{2≥y≥1} trapezoid and triangle together
    https://d6pfkg.bn1302.livefilestore..../a3.png?psid=1
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  7. #22
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    Re: function point

    The symmetry of geometric object

    trapez - y_{x.}=\{|a_{x.}-x_{x.}|\}\cup\{-|a_{x.}-x_{x.}|\} , \{2\geq y_{y.}-2\}\setminus\{1\geq y_{y.}-1\}

    function point-aa1.png


    to make it look a graph a_{x.}\rightarrow a_{y.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{y.})\rightarrow (a_{xy.}x_{xy.})

    a_{x.}\rightarrow a_{z.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{z.})\rightarrow ( x , y , z)
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  8. #23
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    leskovac
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    Re: function point

    a_{x.}\rightarrow a_{y.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{y.})\rightarrow (a_{xy.}x_{xy.})
    function point-12.png
    y_{x.}=|2_{x.}-x_{x.}|
    function point-111.png
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