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anonymous

  • 5 years ago

f(x)=3x^2-3x, f (0)=, f (-1)=, f (2)=

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  1. anonymous
    • 5 years ago
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    plug, then chug.

  2. anonymous
    • 5 years ago
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    what you mean plug than chug

  3. anonymous
    • 5 years ago
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    Replace x with 0 and evaluate the function. Then do the same thing for -1 and 2. If you are sufficiently skilled in math to be evaluating powers (\(x^2\)) you should be sufficiently skilled in substituting values for a variable. Did you have a specific question on how to arrive at an answer here?

  4. anonymous
    • 5 years ago
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    Yes I have never done algebra this is my first year in algebra thats y I said how do you plug and chug

  5. anonymous
    • 5 years ago
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    f(0) means take 0 and replace x with it. Then evaluate the function to find the result.

  6. anonymous
    • 5 years ago
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    \[3(0)^2-3(0) = 3*0 - 3*0 = 0-0 = 0\]

  7. anonymous
    • 5 years ago
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    Now do the same thing for -1 and 2.

  8. anonymous
    • 5 years ago
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    ok so I use the 0 with the -1 and 2

  9. anonymous
    • 5 years ago
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    No. You use the function with -1 and 2.

  10. anonymous
    • 5 years ago
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    I am so sry I am still confused, I guess I just skip over it

  11. anonymous
    • 5 years ago
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    No, don't skip over it. Figure out what is confusing to you. Is English your native language, cause it's not clear to me which part of my explanation doesn't make sense. \[f(x) = 3x^2-3x\] Now if I told you that x = 0 what would you do to find f(x) ?

  12. anonymous
    • 5 years ago
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    will the answer be 0,3,6

  13. anonymous
    • 5 years ago
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    Close.. Should be 0, 6, and 6.

  14. anonymous
    • 5 years ago
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    Yea i didn't do that f(1) right

  15. anonymous
    • 5 years ago
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    it's f(-1)

  16. anonymous
    • 5 years ago
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    right

  17. anonymous
    • 5 years ago
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    At least according to what you wrote initially.

  18. anonymous
    • 5 years ago
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    Thanks so much I think I got it can I be a fan

  19. anonymous
    • 5 years ago
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    So.. \[f(-1) = 3(-1)^2 - 3(-1)\]

  20. anonymous
    • 5 years ago
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    Of course!

  21. anonymous
    • 5 years ago
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    Thanks so much

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spraguer (Moderator)
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