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Math Help - The Graphs of Function and Relation

  1. #1
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    Exclamation The Graphs of Function and Relation

    Graph each equation and identify if it's a mere relation or function by vertical line test.

    1.) |y|=x
    2.) y= x raise to the 3rdpower + 1
    3.) 2y = ( -2x+3 )


    I DON'T know how to answer help pls.
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  2. #2
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    re: The Graphs of Function and Relation

    Quote Originally Posted by ejaykasai View Post
    Graph each equation and identify if it's a mere relation or function by vertical line test.

    1.) |y|=x
    2.) y= x raise to the 3rdpower + 1
    3.) 2y = ( -2x+3 )


    I DON'T know how to answer help pls.
    Dear ejaykasai,

    Let me give you some hints;

    For the first one use the definition of the absolute value function.

    |y| = \begin{cases} y & \mbox{if } y \ge 0  \\ -y  & \mbox{if } y< 0. \end{cases}

    For the second equation use the fist and second derivatives to sketch the graph. If you do not know how to do this please refer Graphing Using First and Second Derivatives

    The third one can be rearranged as, 2y=-2x+3\Rightarrow{y=-x+\frac{3}{2}} So this is a equation of a straight line.

    Hope you can continue.
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  3. #3
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    Re: The Graphs of Function and Relation

    Quote Originally Posted by Sudharaka View Post
    Dear ejaykasai,

    Let me give you some hints;

    For the first one use the definition of the absolute value function.

    |y| = \begin{cases} y & \mbox{if } y \ge 0  \\ -y  & \mbox{if } y< 0. \end{cases}

    For the second equation use the fist and second derivatives to sketch the graph. If you do not know how to do this please refer Graphing Using First and Second Derivatives

    The third one can be rearranged as, 2y=-2x+3\Rightarrow{y=-x+\frac{3}{2}} So this is a equation of a straight line.

    Hope you can continue.
    My problem is that I don't know how to remove the absolute value of y? " | y | "
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  4. #4
    Senior Member BAdhi's Avatar
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    Re: The Graphs of Function and Relation

    What I would simply do is, since I know how to draw the graph of y=|x|, draw it and rotate the whole graph 90 degrees so that y and x axis swap there places
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  5. #5
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    Re: The Graphs of Function and Relation

    Quote Originally Posted by ejaykasai View Post
    My problem is that I don't know how to remove the absolute value of y? " | y | "
    That is exactly what you could do with the definition of the absolute value function.

    |y| = \begin{cases} y & \mbox{if } y \ge 0  \\ -y  & \mbox{if } y< 0. \end{cases}

    |y|=x

    Therefore, y=x\mbox{ when }y\geq 0----------(1)

    -y=x\mbox{ when }y<0\Rightarrow y=-x\mbox{ when }y<0-------(2)

    Combining (1) and (2) we can write;

    y= \begin{cases} x & \mbox{if } y \ge 0  \\ -x  & \mbox{if } y< 0. \end{cases}

    Hope you can sketch the graph now.
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  6. #6
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    Re: The Graphs of Function and Relation

    Quote Originally Posted by BAdhi View Post
    What I would simply do is, since I know how to draw the graph of y=|x|, draw it and rotate the whole graph 90 degrees so that y and x axis swap there places
    Technically, the operation you need to do is reflect the graph about the line y = x. That happens to do the same thing in this case, but in general, the rotation is incorrect. Any time you swap the roles of x and y, you are graphically reflecting about the line y = x.
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  7. #7
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    Unhappy Re: The Graphs of Function and Relation

    Is this right?
    instead of assigning first values of x, values of y 1st, then solve for x:

    Example:

    when y = 0, x = 0 (0, 0)
    when y = 1, x = 1 (1, 1)
    when y = -1, x = 1 (1, -1)
    when y = 2, x = 2 (2, 2)
    when y = -2, x = 2 (2, -2)

    when I solved the |y| = x using the default values, the graph goes like this-->
    |y|=x | Flickr - Photo Sharing! ?? Is this correct?
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  8. #8
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    Re: The Graphs of Function and Relation

    Is this right?
    instead of assigning first values of x, values of y 1st, then solve for x:

    Example:

    when y = 0, x = 0 (0, 0)
    when y = 1, x = 1 (1, 1)
    when y = -1, x = 1 (1, -1)
    when y = 2, x = 2 (2, 2)
    when y = -2, x = 2 (2, -2)

    when I solved the |y| = x using the default values, the graph goes like this-->
    |y|=x | Flickr - Photo Sharing! ?? Is this correct?
    No your graph is incorrect. Please check whether the highlighted values are marked in the graph correctly.
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