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sabika13

  • 3 years ago

The graph of f(x) = ax^4 + bx^2 + cx-24 crosses the x-axis at 1, -2, and 3. determine the equation of f(x).

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  1. anonymous
    • 3 years ago
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    start with \[(x-1)(x+2)(x-3)(x-r)\] and see if you can solve for \(r\)

  2. sabika13
    • 3 years ago
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    do i expand and then solve?

  3. anonymous
    • 3 years ago
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    no you don't really have to expand yet, you can visualize what the constant should be the constant will be \(-6r\) so you know \(-6r=-24\) and if you are lucky \(r=4\) works

  4. anonymous
    • 3 years ago
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    oh damn it doesn't because you get a cube term and your polynomial doesn't seem to have one

  5. anonymous
    • 3 years ago
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    on the other hand something is wrong, because you have a 4th degree polynomial. it has 4 zeros total, although either two or four could be complex. however, if it crosses the \(x\) axis 3 times it must cross it a fourth. if one root is repeated, then it does not cross the axis but only touches it. so i am stumped

  6. sabika13
    • 3 years ago
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    maybe one passes through origin?

  7. anonymous
    • 3 years ago
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    then it still has 4 zeros one would be zero, and also it could not end in \(-24\)

  8. anonymous
    • 3 years ago
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    it really starts with \(ax^4\) right?

  9. sabika13
    • 3 years ago
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    yeah

  10. sabika13
    • 3 years ago
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    i have the answer if that helps?

  11. anonymous
    • 3 years ago
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    ok i am very certain that a fourth degree polynomial cannot cross the \(x\) axis only 3 times

  12. anonymous
    • 3 years ago
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    sure go ahead and let me see why i am wrong

  13. sabika13
    • 3 years ago
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    its f(x)= -4x^3 + 8x^2 + 20 x -20

  14. anonymous
    • 3 years ago
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    we could perhaps start with \[a(x-1)(x+2)(x-3)(x-r)\] and see if we can solve for the coefficients, but it must cross 4 times

  15. sabika13
    • 3 years ago
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    f(x)= -4x^3 + 8x^2 + 20 x -24 *** sorrry

  16. anonymous
    • 3 years ago
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    hold the phone, it is cubic?

  17. sabika13
    • 3 years ago
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    the answers a cubic :S

  18. anonymous
    • 3 years ago
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    i thought it was a 4th degree polynomial, since you wrote \[f(x) = ax^4 + bx^2 + cx-24\]

  19. anonymous
    • 3 years ago
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    if it is \[f(x) = ax^3 + bx^2 + cx-24\] then that is a whole different story

  20. sabika13
    • 3 years ago
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    lol thats what the question says, its ax^4 + bx^2 ..

  21. anonymous
    • 3 years ago
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    and much simpler write \[a(x-1)(x-3)(x+2)= ax^3 + bx^2 + cx-24\] and solve

  22. anonymous
    • 3 years ago
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    then there is something wrong with the question, it is a typo

  23. sabika13
    • 3 years ago
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    so if its ax^3 will the answer be right?

  24. anonymous
    • 3 years ago
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    your question asks for a 4th degree polynomial, which must have 4 zeros. it also has no cube term your answer has a third degree polynomial, so there is a mistake in the question

  25. sabika13
    • 3 years ago
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    yeah it prob is a typo.. ah been trying to question for so long.. thanks tho!

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