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burhan101

  • 2 years ago

find the equation of this graph (picture included) i have started it off but don't know how to finish it

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  1. burhan101
    • 2 years ago
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  2. burhan101
    • 2 years ago
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    \[h(x)=k(x+3)(x-2)(x-3)\]

  3. burhan101
    • 2 years ago
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    3=k(2+3)(2-2)(2-3)

  4. burhan101
    • 2 years ago
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    3=k

  5. CliffSedge
    • 2 years ago
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    Looks alright so far.

  6. burhan101
    • 2 years ago
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    is there more to or is it all ? :S

  7. CliffSedge
    • 2 years ago
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    Er., wait, no

  8. CliffSedge
    • 2 years ago
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    In your solving for k, you need to put in 0 for x if you're using the y-intercept.

  9. burhan101
    • 2 years ago
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    would that make k zero ?

  10. CliffSedge
    • 2 years ago
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    No, what you did made 3=0.

  11. burhan101
    • 2 years ago
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    ohh i see

  12. CliffSedge
    • 2 years ago
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    "3=k(2+3)(2-2)(2-3)" 3=k(5)(0)(-1) -> 3=0

  13. burhan101
    • 2 years ago
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    i cancelled out k too :/

  14. CliffSedge
    • 2 years ago
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    (2,3) is not a solution to the equation, but (0,3) is.

  15. burhan101
    • 2 years ago
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    is dont see where you're getting (2,3) from

  16. CliffSedge
    • 2 years ago
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    You plugged in x=2 and set y=3 in your attempt to solve for k.

  17. CliffSedge
    • 2 years ago
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    So, I don't see where *you* got (2,3) from. ;-)

  18. burhan101
    • 2 years ago
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    ohhhh ! i'm supposed to set x as 0 right ?

  19. CliffSedge
    • 2 years ago
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    Yes, because you know the y-intercept is 3.

  20. burhan101
    • 2 years ago
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    ohh okay i see it now :$

  21. CliffSedge
    • 2 years ago
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    You could use any other known point, but that one is the most convenient.

  22. burhan101
    • 2 years ago
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    so k would be 1/6

  23. burhan101
    • 2 years ago
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    and h(x) = 1/6 (x+3)(x-2)(x-3)

  24. CliffSedge
    • 2 years ago
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    never mind h(x) = k(x+3)(x-2)(x-3) h(0)=3 3=k(3)(-2)(-3) 3=18k

  25. CliffSedge
    • 2 years ago
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    Yep, it's all good.

  26. burhan101
    • 2 years ago
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    Thank-you once again :D

  27. burhan101
    • 2 years ago
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    if you're not busy, can you help me with one last question

  28. CliffSedge
    • 2 years ago
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    I think you know what you're doing, just be careful and don't rush through it.

  29. CliffSedge
    • 2 years ago
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    Oh, sure. I think I can handle one more.

  30. burhan101
    • 2 years ago
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    thanks :)

  31. burhan101
    • 2 years ago
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    Find the equation of a cubic polynomial that gives a remainder of 3, when divided by (x+2)

  32. burhan101
    • 2 years ago
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    how would i approach this problem ?

  33. CliffSedge
    • 2 years ago
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    Do you remember the remainder theorem?

  34. burhan101
    • 2 years ago
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    yes, the factor is equal to zero

  35. CliffSedge
    • 2 years ago
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    Sort of..

  36. CliffSedge
    • 2 years ago
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    It kind of goes like this. If (x+2) is a factor of the polynomial, then dividing by (x+2) would have a remainder of 0. And, if (x+2) is a factor, then x = -2 is a root of the polynomial. If (x+2) is not a factor, then f(-2) = the remainder.

  37. burhan101
    • 2 years ago
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    but how would i use this info to make an equation, this is like working backwards from the norm :/

  38. CliffSedge
    • 2 years ago
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    Yes, a lot of what you're doing here is using concepts to 'unsolve' equations.

  39. CliffSedge
    • 2 years ago
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    If dividing by (x+2) gives a remainder of 3, then you know (-2,3) is a solution to the equation.

  40. burhan101
    • 2 years ago
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    yes that's true

  41. CliffSedge
    • 2 years ago
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    It's a pretty open-ended question; there will be many functions (infinite, really) that you could develop that have that point on it.

  42. burhan101
    • 2 years ago
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    i still don't really get it, like i understand what it is asking for but i don't knw how to get to it :/

  43. CliffSedge
    • 2 years ago
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    It wants a cubic function, so what does an equation for one of those look like?

  44. burhan101
    • 2 years ago
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    like x^3-3x^2-4x-5

  45. CliffSedge
    • 2 years ago
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    Sure, something like that, so to be more general, start with \[y=ax^3+bx^2+cx+d\]

  46. CliffSedge
    • 2 years ago
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    You know what x and y are, and three out of those four variables are free parameters, so choose whatever you want for those, then solve for the fourth one.

  47. CliffSedge
    • 2 years ago
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    You could even set three of them =0 if you wanted to be really lazy about it. (General math tip: be lazy whenever possible)

  48. burhan101
    • 2 years ago
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    so something like \[y=x^3+x^2+x+d \] ?

  49. CliffSedge
    • 2 years ago
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    I don't see why not.

  50. burhan101
    • 2 years ago
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    hmm but "d" has to be a specific number right

  51. CliffSedge
    • 2 years ago
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    Yep, that's what you're solving for. Me, I went with y=ax^3, then solved for a.

  52. CliffSedge
    • 2 years ago
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    (but I'm lazy like that)

  53. burhan101
    • 2 years ago
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    ohh that way a would be -3/8 right

  54. CliffSedge
    • 2 years ago
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    Uh huh.

  55. CliffSedge
    • 2 years ago
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    And if you want, you can verify that \[\large\frac{-3}{8}x^3 \div (x+2)\] will give a remainder of 3.

  56. burhan101
    • 2 years ago
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    \[:. y=\frac{ -3 }{ 8 }x^3\]

  57. CliffSedge
    • 2 years ago
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    Looks beautiful.

  58. burhan101
    • 2 years ago
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    thank you so much for your helppp, you're amazing :D

  59. CliffSedge
    • 2 years ago
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    My pleasure.

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