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anonymous

  • 5 years ago

find a quadratic function given by f(x)=a(x-h)^2+k that models the data 1940,0.25,1968,1.60,1997,5.15 need help solving for a

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  1. amistre64
    • 5 years ago
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    ack!! not this one again..I got a program for this ;)

  2. anonymous
    • 5 years ago
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    lol

  3. amistre64
    • 5 years ago
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    i programmed this just for cases like this

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  4. anonymous
    • 5 years ago
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    can you help me solve for a

  5. amistre64
    • 5 years ago
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    yep

  6. anonymous
    • 5 years ago
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    @amistre64 :D

  7. amistre64
    • 5 years ago
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    0.001301745743669519x^2 -5.0390080805461945x + 4876.675395385017

  8. amistre64
    • 5 years ago
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    :)

  9. amistre64
    • 5 years ago
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    a = something like 64.24/48957 or some rediculaous monster of a fraction

  10. amistre64
    • 5 years ago
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    you wanna know the min wage of 1976 right?

  11. anonymous
    • 5 years ago
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    so when i re write the equation it will look like f(x)= a(x-1940)^2+0.25

  12. anonymous
    • 5 years ago
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    yes

  13. amistre64
    • 5 years ago
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    screenshot

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  14. amistre64
    • 5 years ago
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    1976 = 2.361

  15. amistre64
    • 5 years ago
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    you can write the equation to be a(x-40)^2 to make the numbers easier yes

  16. anonymous
    • 5 years ago
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    so which of the numbers would i put in for a being that i have to show steps

  17. amistre64
    • 5 years ago
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    if we adjust in my program for the -40 we get this: 0.0013017457436695185(x-40)^2 + 0.011765404891539196(x-40) + 0.25

  18. amistre64
    • 5 years ago
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    ive got a post about this same thing a couple of days ago that I typed it all out :)

  19. anonymous
    • 5 years ago
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    thanks

  20. anonymous
    • 5 years ago
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    I'd use y=ax^2+bx+c. They give you values for x and y. Like: 0.25=a(1940)^2+b(1940)+c If you use the three coordinates like that. You will have a system of equations in three variables. Solve for a, b and c. Then complete the square to get it back in vertex form which will probably be a pain.

  21. amistre64
    • 5 years ago
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    id have to redo it all, its a monster tho....here goes :)

  22. amistre64
    • 5 years ago
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    a(1940-1940)^2 + b(1940-1940) + c = .25 from this we see that c = .25 right?

  23. anonymous
    • 5 years ago
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    yes

  24. amistre64
    • 5 years ago
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    im gonna write it as the lesser numbers instead of the years ok? just be easier...

  25. anonymous
    • 5 years ago
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    ok

  26. amistre64
    • 5 years ago
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    the other 2 equations then are: a(28)^2 +b(28) + .25 = 1.60 a(57)^2 +b(57) + .25 = 5.15

  27. amistre64
    • 5 years ago
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    solve for a I suppose is as likely as any other...

  28. amistre64
    • 5 years ago
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    a = 1.60 - .25 -b(28) --------------- now use this value in the other one (28)^2

  29. amistre64
    • 5 years ago
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    write down that value for a, your gonna need it :)

  30. amistre64
    • 5 years ago
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    a(57^2) + b(57) = 5.15 - .25 (57^2)(1.60 - .25 -b(28) ----------------------- + b(57) = 4.60 (28)^2

  31. amistre64
    • 5 years ago
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    (57^2)(1.35) -b(57^2)(28) + b(57)(28^2) = 4.60(28^2)

  32. amistre64
    • 5 years ago
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    b[-(57^2)(28) + (57)(28^2)] = 4.60(28^2) - 57^2(1.35)

  33. amistre64
    • 5 years ago
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    4.60(28^2) - 57^2(1.35) b = ----------------------; thats b lol (57)(28^2) - (57^2)(28)

  34. anonymous
    • 5 years ago
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    wow

  35. amistre64
    • 5 years ago
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    a = 1.60 - .25 -b(28) --------------- ; now plug the value of b into this spot (28)^2

  36. amistre64
    • 5 years ago
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    and thats a

  37. amistre64
    • 5 years ago
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    3606.40 - 4386.15 ----------------- = b 44688 - 90972

  38. amistre64
    • 5 years ago
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    b = 0.016847074583009247256071212514044 if I kept track of it... compare that to the one I posted earlier

  39. amistre64
    • 5 years ago
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    + 0.011765404891539196; this ones better, I probably missed something earlier....

  40. anonymous
    • 5 years ago
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    ok

  41. amistre64
    • 5 years ago
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    but if you must show your work, its just a matter of grunging your way thru it..

  42. amistre64
    • 5 years ago
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    The grunge work is this; formula i worked out ((p1-p2)*((n3*n3)-(n1*n1))+(p3-p1)*((n2*n2)-(n1*n1))) -------------------------------------------------- (((n3*n3)-(n1*n1))*(n1-n2)+((n2*n2)-(n1*n1))*(n3-n1))

  43. amistre64
    • 5 years ago
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    n1 = year1 ; p1 = pay1 and the rest is after this manner

  44. anonymous
    • 5 years ago
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    Lol. Parentheses fun. I always like seeing things solved for variables though.

  45. amistre64
    • 5 years ago
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    \[\frac{[(p1-p2)]*[(n3^2)-(n1^2)]+[(p3-p1)]*[(n2^2)-(n1^2)]}{[(n3^2)-(n1^2)]*[(n1-n2)]+[(n2^2)-(n1^2)]*[(n3-n1)]}\]

  46. amistre64
    • 5 years ago
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    thats 'b' lol

  47. amistre64
    • 5 years ago
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    you also wanna know when it was near $1 right?

  48. amistre64
    • 5 years ago
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    1959 is under a 1 and 1960 is a little over 1

  49. amistre64
    • 5 years ago
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    any questions yet? ;)

  50. amistre64
    • 5 years ago
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    (p2-p1) (n2-n1)*b a = ------------ - ------------- (n2^2)-(n1^2) (n2^2)-(n1^2)

  51. amistre64
    • 5 years ago
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    (p2-p1) - (n2-n1)*b a = ---------------------- is just as good (n2^2)-(n1^2)

  52. anonymous
    • 5 years ago
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    okay so this was the formula that i used to determine when minimum wage was 1.00 is this correct y=1.00(1964-1940)^2+.25 y=1.00*234^2+.25 y=(1.00*.576)+.25 y=.58+.25=1.08

  53. amistre64
    • 5 years ago
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    no, thats not correct minwage = y = 1 $1 is not a constant for the equation; it is a solution to it

  54. amistre64
    • 5 years ago
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    and you are missing your middle term

  55. amistre64
    • 5 years ago
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    you would take $1 from c(.25) setting the whole equation = 0; then use the quadratic formula to calculate roots as solutions and throw out the obvious wrong one

  56. anonymous
    • 5 years ago
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    so could you show me how to write the equation and i try to solve it again

  57. amistre64
    • 5 years ago
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    or trial and error it like I did ;)

  58. amistre64
    • 5 years ago
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    lets do this the generic way to get you to the formula I alrady created; and then all you need to do is fill in the plugs.... ok? thatll save a whole lot of little numbers getting orphaned :)

  59. anonymous
    • 5 years ago
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    ok

  60. amistre64
    • 5 years ago
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    y = year and p = price; makes sense? a(y1)^2 + b(y1) + c = (p1)

  61. amistre64
    • 5 years ago
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    a(y1)^2 + b(y1) + c = (p1) Eq1 a(y2)^2 + b(y2) + c = (p2) Eq2 a(y3)^2 + b(y3) + c = (p3) Eq3

  62. amistre64
    • 5 years ago
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    eq1; c = (p1) -b(y1) - a(y1)^2

  63. amistre64
    • 5 years ago
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    laptop cord got unplugged lol

  64. amistre64
    • 5 years ago
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    Eq ; a(y2)^2 + b(y2) + c = (p2) a(y2)^2 + b(y2) + (p1) -b(y1) - a(y1)^2 = (p2) [a(y2)^2 - a(y1)^2] + [b(y2) -b(y1)] = (p2) - (p1) a[(y2)^2 -(y1)^2] + b[(y2) -(y1)] = (p2) - (p1) a = (p2) - (p1) - [(y2) -(y1)]b --------------------- [(y2)^2 -(y1)^2] thats what a equals a[(y2)^2 -(y1)^2] = (p2) - (p1) - [(y2) -(y1)]b

  65. amistre64
    • 5 years ago
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    forget that last part, its seems a bit spurious.... forgot to delete it lol

  66. amistre64
    • 5 years ago
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    From Eq3 we substitute all the values if for a and c a(y3)^2 + b(y3) + c = (p3) (y3^2) [(p2) - (p1) - [(y2) -(y1)]b ---------------------------- [(y2)^2 -(y1)^2] + b(y3) + (p1) -b(y1) - [(y1)^2](y3^2) [(p2) - (p1) - [(y2) -(y1)]b --------------------------------- [(y2)^2 -(y1)^2]

  67. amistre64
    • 5 years ago
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    we end up solving this for b and the plugging in the values that I already postd for B above :)

  68. amistre64
    • 5 years ago
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    you feel up to all that?

  69. anonymous
    • 5 years ago
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    have to be up to it lol

  70. amistre64
    • 5 years ago
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    I posted all the formulas; start with b go to a and then...well c is .25 for this one lol

  71. amistre64
    • 5 years ago
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    B = (.25-1.60) [(57^2)-(28^2)]+(5.15-.25)[(28^2)-(0^2)] ----------------------------------------------- (0 -28)[(57^2)-(28^2)]+(57-28)[(28^2)-(0^2)]

  72. anonymous
    • 5 years ago
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    ok this is how i did it lol(b) I believe that the quadratic function that models the data is y = a (y - 1940)2 + .25 because it seems right. (c) Estimate the minimum wage in 1976 and compare it to the actual value of $2.30. y = $1.99 (1976 – 1940)2 + .25 y = $1.99 · 362 + .25 y = ($1.99 · 1296) + .25 y = $2.58 + .25 = $2.83 Estimated $2.83 (d) Estimate when the minimum wage was $1.00. y = $1.00 (1964 – 1940)2 + .25 y = $1.00 · 242 + .25 y = ($1.00 · 576) + .25 y = $.58 + .25 = $1.08 The year 1964 (e) If current trends continue, predict the minimum wage in 2009. Compare it to the projected value of $7.25. y = $7.25 (2009 – 1940)2 + .25 y = $7.25 · 692 + .25 y = ($7.25 · 4761) + .25 y = $3.45 + .25 = $3.70 Minimum wage $3.70 @ a $3.55 difference

  73. anonymous
    • 5 years ago
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    i might have jacked it up sorry

  74. amistre64
    • 5 years ago
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    If it works for you use it ;) but dont use y for year and y for a dependant variable, its confusing lol

  75. amistre64
    • 5 years ago
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    the way you solve for it depends on the degree of accuracy you are looking for.

  76. anonymous
    • 5 years ago
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    so take y out? do i do them correct

  77. anonymous
    • 5 years ago
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    did i do them correct

  78. amistre64
    • 5 years ago
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    I dont know how you did them ;) but you shouldnt use y to mean year AND y to mean value of the function....

  79. anonymous
    • 5 years ago
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    ok

  80. amistre64
    • 5 years ago
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    my way gives me 7.259 for min wage in 2007

  81. amistre64
    • 5 years ago
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    compared to 7.25 thats pretty accurate

  82. amistre64
    • 5 years ago
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    1960 gives me 1.006

  83. anonymous
    • 5 years ago
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    okay so i re did equation F (0) = .25 (0 -1.60)^2 + 5.15 .25(-1.60)^2 + 5.15 .25(2.56) + 5.15 .64 +5.15

  84. amistre64
    • 5 years ago
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    i dont understand the process of how you get your equation; im sure it has to do with homogeneous difference equattions tho...

  85. amistre64
    • 5 years ago
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    the only way I know how to solce for quadratics with three data points is to plug them into the quad equation and solve for a b and c

  86. amistre64
    • 5 years ago
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    and that aint what you did there :)

  87. anonymous
    • 5 years ago
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    okay lets scratch last equation lol

  88. amistre64
    • 5 years ago
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    do you know how to solve for a system of equations?

  89. amistre64
    • 5 years ago
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    c= .25 a(28^2) + b(28) + c = 1.60 a(58^2) + b(58) + c = 5.15 solve by substitution, elimination or matrix....

  90. amistre64
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    a(28^2) + b(28) = 1.60-.25 = 1.35 a(58^2) + b(58) = 5.15-.25 = 4.90 a(28^2) + b(28) = 1.35 a = [1.35 - b(28)]/(28^2) ...................................... a(58^2) + b(58) = 4.90 ; plug in value for a

  91. amistre64
    • 5 years ago
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    (58^2)[1.35 - b(28)] + b(58)(28^2) = 4.90(28^2) 58^2(1.35) -b(58^2)(28) + b(58)(28^2) = 4.90(28^2) b[(-58^2)(28) +(58)(28^2)] = 4.90(28^2) - 58^2(1.35) b = 4.90(28^2) - 58^2(1.35) --------------------- (-58^2)(28) +(58)(28^2)

  92. amistre64
    • 5 years ago
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    plug those numbers into a calculator to get the solid value for 'b' and use it in the equation above it to solve for 'a'

  93. amistre64
    • 5 years ago
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    ((4.90 * (28^2)) - ((58^2) * 1.35)) / (((-(58^2)) * 28) + (58 * (28^2))) = 0.014363711 Google says b = .014363711

  94. amistre64
    • 5 years ago
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    a = [1.35 - b(28)]/(28^2) (1.35 - (.014363711 * 28)) / (28^2) = 0.0012089491

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