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Math Help - Graphing

  1. #1
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    Graphing

    Graph the following equation:

    (2x+5)^1/3 +3=0

    x + 1 =3
    x^2-4 x+2
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  2. #2
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    Graph the following equation:

    (2x+5)^1/3 +3=0

    x + 1 =3
    x^2-4 x+2
    There is a solution to this equation, but the graph of that solution will be a straight vertical line (not a function) -- are you sure you are not supposed to graph the function (2x + 5)^(1/3) + 3 = y?
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  3. #3
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    This was part of the question. the entire question was 1) find all solutions of the equation algebraically 2) then, check your solution both algebraically and graphically.


    I understood how to do the first part but i didn't know how to graph it (2nd part)?
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  4. #4
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    This was part of the question. the entire question was 1) find all solutions of the equation algebraically 2) then, check your solution both algebraically and graphically.


    I understood how to do the first part but i didn't know how to graph it (2nd part)?

    Ok, so first things first -- lets make sure you got the right answer to the equation. What did you come up with?
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  5. #5
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    This was part of the question. the entire question was 1) find all solutions of the equation algebraically 2) then, check your solution both algebraically and graphically.


    I understood how to do the first part but i didn't know how to graph it (2nd part)?
    The next step, assuming you have the correct solution, is to graph not the solution but the original function -- y = (2x + 5)^(1/3) + 3 -- and check that the ordered pair (x,0) on the graph matches your solution. You don't need to solve anything to graph the function, since it is already in the form f(x)=x. There are two ways to go about graphing it: transformations of the graph of
    f(x) = x^(1/3), which is the more formal way; or by just plugging in values for x and finding the corresponding values for y and drawing a smooth curve between those points.
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  6. #6
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    My solution was :

    x=-16
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  7. #7
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    My solution was :

    x=-16

    I am not quite sure what you mean in the second part of your answer. Sorry
    Ok, you have the right solution, and that is the place to start! Have you learned about parent graphs and transformations yet?
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  8. #8
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    somewhat.. we really didnt focus much on cube root transformations
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  9. #9
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    somewhat.. we really didnt focus much on cube root transformations
    Well, the principles of transformation apply to any graph, but that is still sort of the involved way to do this if you are not that familiar with transformations. In order to draw a graph, all you really need is a set of ordered pairs (x,y) for which the equation is true. You find those by solving the equation for a series of values of x over the range where you want to draw the graph.
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  10. #10
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    ohh so i just need to plug in values for x and y?


    and also there was another equation below the one we have been doing at the top. it is of different format but maybe you could help me

    thanks!!!
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  11. #11
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    ohh so i just need to plug in values for x and y?


    and also there was another equation below the one we have been doing at the top. it is of different format but maybe you could help me

    thanks!!!
    Plug in values for x and solve for y, but yes, that will generate a set of ordered pairs you can use to draw the graph. It helps if you know the general shape you are going for, so if you have a graphing calculator you can plug in the equation to see the general shape. And yes, I can help you with the next one.
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  12. #12
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    thank you!!!!
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  13. #13
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    thank you!!!!
    Your welcome! I will be a little while (about an hour) before I can get back to you on the next one, but will be happy to help.
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  14. #14
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    no problem :]
    as long as i understand it, i am good.
    once again thanks!
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