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heathernelly

kkkkk

  • one year ago
  • one year ago

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  1. dpaInc
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    can you give me the derivative of f ??

    • one year ago
  2. heathernelly
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    that's all it says :// :(

    • one year ago
  3. dpaInc
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    yes... but you need to calculate the derivative of f to find the relative max/min of the function.

    • one year ago
  4. dpaInc
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    isn't this for a calculus class?

    • one year ago
  5. heathernelly
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    algebra 2

    • one year ago
  6. dpaInc
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    oh... sorry....

    • one year ago
  7. heathernelly
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    its okay :) so you don't know how to do it? :(

    • one year ago
  8. heathernelly
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    i know how to do it on a calculater ..but i left it at a friends house ://

    • one year ago
  9. dpaInc
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    oh... so you're allowed to use the max/min functions of a graphing calculator?

    • one year ago
  10. heathernelly
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    yes

    • one year ago
  11. dpaInc
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    ok... hang on...

    • one year ago
  12. heathernelly
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    okaay :)

    • one year ago
  13. dpaInc
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    since you don't have your graphing calculator with you, use this online one.... https://www.desmos.com/calculator i've also included the graph of your function here:

    • one year ago
  14. dpaInc
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    it looks like you have a relative max at x=-4 and a relative min at x=0

    • one year ago
  15. heathernelly
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    yeah, but that's not one of my options :(

    • one year ago
  16. dpaInc
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    the function you gave at the start does not coincide with those choices...

    • one year ago
  17. heathernelly
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    thats what it says in my homework

    • one year ago
  18. dpaInc
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    i double checked with this function you gave at the start: \(\large f(x)=x^3+6x^2-36 \) f has a relative max at x=-4 and a relative min at x=0.

    • one year ago
  19. dpaInc
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    f(-4) = -4 so (-4, -4) is a relmax f(0) = -36 so (0, -36) is a relmin

    • one year ago
  20. heathernelly
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    but that's none of the options ://

    • one year ago
  21. dpaInc
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    i'm sorry.... i'm absolutely sure of my answer here.. even the calculator verifies my answer.... is this the function we started with correct?

    • one year ago
  22. heathernelly
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    yes it is ..hmm :// :(

    • one year ago
  23. heathernelly
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    im just going to pick a or b:///.. lol ..thank you for the help though :)

    • one year ago
  24. dpaInc
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    idk what to say... :(

    • one year ago
  25. heathernelly
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    do you think you could help me with another one?

    • one year ago
  26. dpaInc
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    there has to be an error in those choices... yw... :)

    • one year ago
  27. dpaInc
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    yes...

    • one year ago
  28. heathernelly
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    What is a cubic polynomial function in standard form with zeros 1, –2, and 2?

    • one year ago
  29. dpaInc
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    do you know how to multiply out binomials?

    • one year ago
  30. heathernelly
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    i know i have to write like (x+1)(x-2)(x+2)

    • one year ago
  31. dpaInc
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    standard form for a cubic polynomial is: \(\large y=ax^3+bx^2+cx+d \) where a, b, c, d are real numbers.

    • one year ago
  32. dpaInc
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    not quite... since the zeros are 1, -2, and 2, then the factors are (x-1)(x+2)(x-2)

    • one year ago
  33. dpaInc
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    so all you do is multiply out the 3 binomials....

    • one year ago
  34. heathernelly
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    how is it -1 when it says 1 ?

    • one year ago
  35. dpaInc
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    a zero refers to where the graph crosses the x-axis (when y=0) so the equation is 0 = (x-1)(x+2)(x-2) if x=1, you'll have 0 = (1-1)(1+2)(1-2) which is a true statement but if you have what you wrote down, 0 = (1+1)(1+2)(1-2) is not a true statement

    • one year ago
  36. dpaInc
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    multiply out the binomials: (x-1)(x+2)(x-2) = \(\large x^3-x^2-4x+4 \)

    • one year ago
  37. heathernelly
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    thanks :)

    • one year ago
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