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heathernelly

  • 2 years ago

kkkkk

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  1. dpaInc
    • 2 years ago
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    can you give me the derivative of f ??

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

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

  4. dpaInc
    • 2 years ago
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    isn't this for a calculus class?

  5. heathernelly
    • 2 years ago
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    algebra 2

  6. dpaInc
    • 2 years ago
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    oh... sorry....

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

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

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

  10. heathernelly
    • 2 years ago
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    yes

  11. dpaInc
    • 2 years ago
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    ok... hang on...

  12. heathernelly
    • 2 years ago
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    okaay :)

  13. dpaInc
    • 2 years ago
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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:

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

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

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

  17. heathernelly
    • 2 years ago
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    thats what it says in my homework

  18. dpaInc
    • 2 years ago
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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.

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

  20. heathernelly
    • 2 years ago
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    but that's none of the options ://

  21. dpaInc
    • 2 years ago
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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?

  22. heathernelly
    • 2 years ago
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    yes it is ..hmm :// :(

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

  24. dpaInc
    • 2 years ago
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    idk what to say... :(

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

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

  27. dpaInc
    • 2 years ago
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    yes...

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

  29. dpaInc
    • 2 years ago
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    do you know how to multiply out binomials?

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

  31. dpaInc
    • 2 years ago
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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.

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

  33. dpaInc
    • 2 years ago
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    so all you do is multiply out the 3 binomials....

  34. heathernelly
    • 2 years ago
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    how is it -1 when it says 1 ?

  35. dpaInc
    • 2 years ago
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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

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

  37. heathernelly
    • 2 years ago
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    thanks :)

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