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AutumnMaybury

  • 3 years ago

if f(x)=x^2-2x+4, find f(-3x)

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  1. hba
    • 3 years ago
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    Just plug in -3x in place of x :)

  2. hba
    • 3 years ago
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    @AutumnMaybury Start working :)

  3. AutumnMaybury
    • 3 years ago
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    answer is one?

  4. hba
    • 3 years ago
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    @AutumnMaybury What ?

  5. AutumnMaybury
    • 3 years ago
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    i got one for the answer

  6. hba
    • 3 years ago
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    Please show me your working @AutumnMaybury

  7. AutumnMaybury
    • 3 years ago
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    f(x)=-3^2-2(-3)+4

  8. hba
    • 3 years ago
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    Its not -3,It's -3x

  9. hba
    • 3 years ago
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    See the question you provided us

  10. AutumnMaybury
    • 3 years ago
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    but you said put -3 in for x..... *confused*

  11. hba
    • 3 years ago
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    I said put '-3x' in place of x

  12. AutumnMaybury
    • 3 years ago
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    ok sorry rewriting now

  13. hba
    • 3 years ago
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    Ok :)

  14. AutumnMaybury
    • 3 years ago
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    3x+4

  15. hba
    • 3 years ago
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    @AutumnMaybury I want you to show me your working please :)

  16. AutumnMaybury
    • 3 years ago
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    |dw:1357830860437:dw|

  17. hba
    • 3 years ago
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    Well lemme show you something \[\huge\ (-3x)^2=(-3)^2(x)^2=9x^2\]

  18. AutumnMaybury
    • 3 years ago
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    is that the answer?? please explain *confused* :(

  19. hba
    • 3 years ago
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    It would be x^2 not x

  20. hba
    • 3 years ago
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    I mean to say the first one my bad.

  21. hba
    • 3 years ago
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    f(x)=9x^2-6x+4

  22. AutumnMaybury
    • 3 years ago
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    but 9x-6x does equal 3x-4

  23. hba
    • 3 years ago
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    It would be 9x^2 not 9x.

  24. AutumnMaybury
    • 3 years ago
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    is the 9 the only one squareD? the 6x isnt?

  25. hba
    • 3 years ago
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    No.

  26. hba
    • 3 years ago
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    The 6x is not.

  27. hba
    • 3 years ago
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    I have showed you above in latex.

  28. AutumnMaybury
    • 3 years ago
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    ok thank you

  29. hba
    • 3 years ago
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    |dw:1357831302413:dw|

  30. AutumnMaybury
    • 3 years ago
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    closing out of this question anddddd askingg a neww onee

  31. hba
    • 3 years ago
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    Okay :)

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