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Math Help - quadratic equations

  1. #1
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    quadratic equations

    hi, i need to make this into a quadrattic equation but there is a pert on the table which i can't remember what to do!

    when you have x then -3,-2,-1,0,1,2,3
    under that you puy Y and i can remember how to work that out! if you can help i will be very thankful!

    my question is Y= x^2-4x+1 by the way
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Replace the values of x in the equation given to you!

    For example, under -3, you put:

    (-3)^2 - 4(-3) + 1 = 9 + 12 + 1 = 22

    Hence:

    \begin{array}{|c|c|c|c|c|c|c|c|}\hline<br />
x & -3 & -2 & -1&0&1&2&3 \\ \hline<br />
y & 22 &&&&&&\\ \hline \end{array}
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  3. #3
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    thanks for answering my question! under -2 it would be 11?
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  4. #4
    MHF Contributor Unknown008's Avatar
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        (-2)^2 - 4(-2) + 1 = 4 + 8 + 1

    Does this equal 11?
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  5. #5
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    no its 13, i think i got it now!! would -1=3?
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  6. #6
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    Quote Originally Posted by andyboy179 View Post
    would -1=3?
    I get 6
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Think of it again:

         (-1)^2 - 4(-1) + 1 = 1 + 4 + 1

    Does this make 1?

    remember that when you multiply two negative numbers, they become a positive number.
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  8. #8
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    would 0 =1
    1=0
    2=1
    3=1???
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  9. #9
    MHF Contributor Unknown008's Avatar
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    for 0: yes.

    For 1:      (1)^2 - 4(1) + 1 = 1 - 4 + 1

    For 2:      (2)^2 - 4(2) + 1 = 4 - 8 + 1

    For 3:      (3)^2 - 4(3) + 1 = 9 - 12 + 1

    Hm.. you need to work your basic calculations more...
    Last edited by Unknown008; October 22nd 2010 at 09:11 AM. Reason: Typo. Changed "For 2:" to "For 3:"
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  10. #10
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    actually would this be it:
    -3=19
    -2=11
    -1=5
    0=1
    1=-1
    2=-1
    3=1
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  11. #11
    MHF Contributor Unknown008's Avatar
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    Are you sure you posted the right equation?
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  12. #12
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    cr@p, no i didn't its -4X sorry!!!!!!!!
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  13. #13
    MHF Contributor Unknown008's Avatar
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    I still have no idea what the equation actually is...
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