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Math Help - Function and Relation Help!

  1. #1
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    Function and Relation Help!

    Please help me with this problem. How can I form equations out of the given relations and graph each?

    Here is the given relations

    a. (2, 1), (1, -1), (-1, -5), (-2, -7)
    b. (1, -2), (2, -4), (-1, 2)

    Example:

    (4,8), (2,4), (-4,-8), (-1,-2), (3,6)
    Answer: y = 2x

    (3,1), (-1,-3), (0,-2), (2,0), (1,-1)
    Answer: y = x-2
    ___

    Is there any easy way where I can find the equation of a given relation? Because based on the examples, it's quite easy but in the exercise the given relations are hard.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Hi888 View Post
    Please help me with this problem. How can I form equations out of the given relations and graph each?

    Here is the given relations

    a. (2, 1), (1, -1), (-1, -5), (-2, -7)
    b. (1, -2), (2, -4), (-1, 2)

    Example:

    (4,8), (2,4), (-4,-8), (-1,-2), (3,6)
    Answer: y = 2x

    (3,1), (-1,-3), (0,-2), (2,0), (1,-1)
    Answer: y = x-2
    ___

    Is there any easy way where I can find the equation of a given relation? Because based on the examples, it's quite easy but in the exercise the given relations are hard.
    For this kind of things, you have to do trials & errors.

    For example b) :
    you can "see" that there is a common ratio between the 2 numbers :

    (-2)\bold{1}=\bold{-2}

    (-2)\bold{2}=\bold{-4}

    etc...



    For a) :
    You have no idea how to do. Because you know what the examples look like, see if it is like in b), where there is a common ratio :
    Is \frac 21=\frac 1{-1} ?
    Answer is no.

    Now try common difference.

    Is 2-1=1-(-1)=-1-(-5) ?
    Aswer is no.


    So now, let's try something in the form ax+b.

    Is there a unique solution to the system \left\{\begin{array}{ccc} 1&=&a*2+b \\ -1&=&a*(-1)+b \end{array} \right. ?

    See if the solutions you find fit with the other couples





    This is my 24th post ^^
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  3. #3
    Senior Member nikhil's Avatar
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    Check it out

    Equation for relation a will be
    2x-y=3
    equation for b will be
    y=-2x
    There is only one method to solve this that is observation.
    There might be relations that may be satisfied by more than 1 equation.In relations that u gave had linear relation (just observed) which always make a straight line on graph. So by using the relation an equation of straight line was made.but there may be any other equation which may not be linear but may also satisfy the given relation(happens when two graphs intersect,at intersection point both show same value though overall behaviour may be different.
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  4. #4
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    Hello, Hi888!

    How can I form equations out of the given relations and graph each?

    Here are the given relations

    a. (2, 1), (1, -1), (-1, -5), (-2, -7)

    b. (1, -2), (2, -4), (-1, 2)


    Examples:

    (4, 8), (2, 4), (-4, -8), (-1, -2), (3, 6)
    . . Answer: y \:= \:2x

    (3, 1), (-1, -3), (0, -2), (2, 0), (1, -1)
    . . Answer: y \:=\: x-2
    I would write the pairs in increasing values of x.


    Let's look at the first example . . .

    . . \begin{array}{c|c|c|c|c|c}\hline x & \text{-}4 & \text{-}1 & 2 & 3 & 4 \\ \hline y & \text{-}8 & \text{-}2 & 4 & 6 & 8 \\ \hline \end{array}

    Upon inspection, we see that y is always twice x.

    Therefore: . y \:=\:2x



    Look at the second example . . .

    . . \begin{array}{c|c|c|c|c|c}\hline x & \text{-}1 & 0 & 1 & 2 & 3 \\ \hline y & \text{-}3 & \text{-}2 & \text{-}1 & 0 & 1 \\ \hline \end{array}

    We see that y is always two less than x.

    Therefore: . y \:=\:x-2



    Now consider problem (b): . \begin{array}{c|c|c|c|c}\hline x & \text{-}1 & 0 & 1 & 2 \\ \hline y & 2 & ? & \text{-}2 & \text{-}4 \\ \hline \end{array}

    We see that y is always -2 times x.

    Therefore: . y \:=\:-2x



    Problem (a) is trickier: . \begin{array}{c|c|c|c|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -7 & -5 & ? & -1 & 1 \\ \hline\end{array}

    Note that y "goes up by 2's."
    . . This means that the function contains 2x.

    Make a new chart: . \begin{array}{c|c|c|c|c|c}<br />
x & \text{-}2 & \text{-}1 & 0 & 1 & 2 \\ \hline<br />
2x & \text{-}4 & \text{-}2 & 0 & 2 & 4 \\ \hline<br />
y & \text{-}7 & \text{-}5 & ? & \text{-}1 & 1 \\ \hline \end{array}

    We see that y is 3 less than 2x.

    Therefore: . y \:=\:2x-3



    Yes, this is a very primtive approach to these problems.
    . . But it is convenient for simple (linear) functions.

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