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anonymous

  • 5 years ago

if f(x)=3x^2-2x-5, find(2u+1) can you explain how to work the problem and whats the answer i can get to f(2u+1)=3(2u+1) after that i don't know

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  1. radar
    • 5 years ago
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    \[f(2u+1)=3(2u+1)^{2}-2(2u+1)-5\] Thats a start.

  2. radar
    • 5 years ago
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    \[=3(4u ^{2}+4u+1)-4u-2-5\] just keep on trucking!

  3. radar
    • 5 years ago
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    Earth to shrryflores, do you follow so far?

  4. anonymous
    • 5 years ago
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    yes

  5. radar
    • 5 years ago
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    Let's continue then: \[12u ^{2}+12u+3-4u-7\] \[12u ^{2}+8u+4\]

  6. radar
    • 5 years ago
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    change that +4 to -4

  7. anonymous
    • 5 years ago
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    ok it makes since

  8. radar
    • 5 years ago
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    since there was no = sign I think that is as far as needed\[12u ^{2}+12u-4\]

  9. radar
    • 5 years ago
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    \[f(2u+1)=12u ^{2}+12u-4\]

  10. anonymous
    • 5 years ago
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    \[f(2u+1)=12u^2+8u-4\] You can find the roots of u using the quadratic formula: \[u=-\frac{1}{3}\pm \frac{2}{3}\rightarrow u=-1, u=\frac{1}{3}\]

  11. radar
    • 5 years ago
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    I was wondering if it was to be taken any further? Couldn't determine if they were wanting the roots.

  12. anonymous
    • 5 years ago
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    I gave the roots just in case

  13. radar
    • 5 years ago
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    Fine business, they say "whats the answer" maybe that was what they wanted. thanks and good luck to you and shrryflores

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