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anonymous

  • one year ago

Without plotting any points other than intercepts, draw a possible graph of the following polynomial: f(x) = (x + 8)^3(x + 6)^2(x + 2)(x − 1)3(x − 3)^4(x − 6).

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    is your function like this: \[{\left( {x + 8} \right)^3}{\left( {x + 6} \right)^2}\left( {x + 2} \right){\left( {x - 1} \right)^3}{\left( {x - 3} \right)^4}\left( {x - 6} \right)\]

  3. anonymous
    • one year ago
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    Yes :)

  4. Michele_Laino
    • one year ago
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    the graph of your function crosses the x-axis at points x=-8, x=-6, x=-2, x=1, x=3, and x=6 since at those points your function: \[f\left( x \right) = {\left( {x + 8} \right)^3}{\left( {x + 6} \right)^2}\left( {x + 2} \right){\left( {x - 1} \right)^3}{\left( {x - 3} \right)^4}\left( {x - 6} \right)\] is equal to zero

  5. Michele_Laino
    • one year ago
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    |dw:1436971271054:dw|

  6. Michele_Laino
    • one year ago
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    now, we have this: \[f\left( { - 9} \right) = {\left( { - 9 + 8} \right)^3}{\left( { - 9 + 6} \right)^2}\left( { - 9 + 2} \right){\left( { - 9 - 1} \right)^3}{\left( { - 9 - 3} \right)^4}\left( { - 9 - 6} \right) > 0\]

  7. Michele_Laino
    • one year ago
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    whereas: \[f\left( 7 \right) = {\left( {7 + 8} \right)^3}{\left( {7 + 6} \right)^2}\left( {7 + 2} \right){\left( {7 - 1} \right)^3}{\left( {7 - 3} \right)^4}\left( {7 - 6} \right) > 0\]

  8. Michele_Laino
    • one year ago
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    so the graph of your function can be like this: |dw:1436971518126:dw|

  9. anonymous
    • one year ago
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    Thanks:) You're a LIFESAVER!!!

  10. Michele_Laino
    • one year ago
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    thanks!! :)

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