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IsTim

Determine an equation for the tangent to each curve at the indicated value. P(x)=(-x^3+2)(4x^2-3),x=3

  • one year ago
  • one year ago

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  1. IsTim
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    I've already completed 96% of the question. My final answer is y=-149x+4481, but the textbook says it is y=-1491x+3648

    • one year ago
  2. Mimi_x3
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    I got \(y=-1491x+3648\) as well..probably you did something wrong

    • one year ago
  3. IsTim
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    Er...what did you get as your y-value?

    • one year ago
  4. Mimi_x3
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    \[y-y_1 =m(x-x_1)\] \[y+825=-1491(x-3) => y+825=-1491x+4473 => y-3648=-1491x\] \[=> y =-1491x+3648\]

    • one year ago
  5. Mimi_x3
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    hm, want me to explain? or you understand it?

    • one year ago
  6. IsTim
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    How did you get your y-value? I did this:\[y=(-(3)^{3}+2)(4(3)^{2}-3)\]

    • one year ago
  7. Mimi_x3
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    The minus is inside the bracket.

    • one year ago
  8. Mimi_x3
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    \[y=((-3)^3+2)(4(3)^2-3)\]

    • one year ago
  9. Mimi_x3
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    then the y-valule is \(-825\)

    • one year ago
  10. IsTim
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    No, I don't think the minus sign is inside the bracket... My equation is this actually, sorry. \[y=(−(3)^{3}+2)(4(3)^{2}−3)\] I'll try what you said though.

    • one year ago
  11. Mimi_x3
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    Hmm..is it \[P(x) = (-x^3+2)(4x^2-3)\] ?

    • one year ago
  12. IsTim
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    Ignore what I said. Thanks. I got it.

    • one year ago
  13. Mimi_x3
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    (:

    • one year ago
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