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anonymous

  • 5 years ago

can someone help me find the critical points for: .288x2 +3.47x +5.46?

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  1. amistre64
    • 5 years ago
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    f' = .288(2)x +3.47 solve for f'=0 x = -3.47/(.288(2))

  2. amistre64
    • 5 years ago
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    actually its even simpler than that, only one critical point exists because this graph looks like a "U". The critical point is at the bottom part of th "U"

  3. anonymous
    • 5 years ago
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    its not multiplied by 2 thats supposed to be a power...sorry typo

  4. amistre64
    • 5 years ago
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    -3.47/[2(.288)] comes out the same, but you aint gotta try to derive it :)

  5. amistre64
    • 5 years ago
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    right, I assumed that

  6. anonymous
    • 5 years ago
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    is a critical point a coordinate?

  7. amistre64
    • 5 years ago
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    6.024 is pretty close, it is the "x" coordinate; just plug it into the equation to get the "y" coordinate.

  8. anonymous
    • 5 years ago
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    the x coordinate is -6, so I plug this in to the original equation?

  9. amistre64
    • 5 years ago
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    (6.024,36.81) is what I get for a coordinate if I did it right :)

  10. amistre64
    • 5 years ago
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    yeah.... -6... which means I plugged it in wring ...

  11. anonymous
    • 5 years ago
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    wait can you tell me how you did that?

  12. amistre64
    • 5 years ago
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    which part?

  13. amistre64
    • 5 years ago
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    do you know how to find a derivative?

  14. anonymous
    • 5 years ago
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    how you find the coordinate? and is that coordinate the critical point?

  15. amistre64
    • 5 years ago
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    to find which value of x gets us a critical point, we can find the derivative of an equation if you do not know what the graph of the equation will look like

  16. amistre64
    • 5 years ago
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    y =.288x^2 +3.47x + 5.46 the derivative is gotten thru using the rules for derivatives, mainly the power rule for this equation.

  17. amistre64
    • 5 years ago
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    the power rule states: when ever you have a variable to the power of a number like: x^4 you can multiply the power by whatever number is in front of the "x" and then reduce the power by 1 (4) x ^ 4-1 = 4x^3 does that make sense?

  18. anonymous
    • 5 years ago
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    yes...then what?

  19. amistre64
    • 5 years ago
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    y =.288x^2 +3.47x + 5.46 y' = (2).288x^1 + (1)3.47x^0 y' = .165888 x + 3.47 when y' = 0 we have found a critical point for "x". so we set y' = 0 0 = .165888x + 3.47 solve for "x" -3.47/.165888 = x this is the value of x that we need to use in the original equation; the y = .288x^2 +3.47x +5.46

  20. amistre64
    • 5 years ago
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    and I messed up that 2(.288).... redo that will ya :)

  21. amistre64
    • 5 years ago
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    x = -3.47/.576 which is about -6

  22. anonymous
    • 5 years ago
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    ok thanks! could the critical point and the minimum point be the same?

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