## 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

1. amistre64

ack!! not this one again..I got a program for this ;)

2. anonymous

lol

3. amistre64

i programmed this just for cases like this

4. anonymous

can you help me solve for a

5. amistre64

yep

6. anonymous

@amistre64 :D

7. amistre64

0.001301745743669519x^2 -5.0390080805461945x + 4876.675395385017

8. amistre64

:)

9. amistre64

a = something like 64.24/48957 or some rediculaous monster of a fraction

10. amistre64

you wanna know the min wage of 1976 right?

11. anonymous

so when i re write the equation it will look like f(x)= a(x-1940)^2+0.25

12. anonymous

yes

13. amistre64

screenshot

14. amistre64

1976 = 2.361

15. amistre64

you can write the equation to be a(x-40)^2 to make the numbers easier yes

16. anonymous

so which of the numbers would i put in for a being that i have to show steps

17. amistre64

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

ive got a post about this same thing a couple of days ago that I typed it all out :)

19. anonymous

thanks

20. anonymous

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

id have to redo it all, its a monster tho....here goes :)

22. amistre64

a(1940-1940)^2 + b(1940-1940) + c = .25 from this we see that c = .25 right?

23. anonymous

yes

24. amistre64

im gonna write it as the lesser numbers instead of the years ok? just be easier...

25. anonymous

ok

26. amistre64

the other 2 equations then are: a(28)^2 +b(28) + .25 = 1.60 a(57)^2 +b(57) + .25 = 5.15

27. amistre64

solve for a I suppose is as likely as any other...

28. amistre64

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

29. amistre64

write down that value for a, your gonna need it :)

30. amistre64

a(57^2) + b(57) = 5.15 - .25 (57^2)(1.60 - .25 -b(28) ----------------------- + b(57) = 4.60 (28)^2

31. amistre64

(57^2)(1.35) -b(57^2)(28) + b(57)(28^2) = 4.60(28^2)

32. amistre64

b[-(57^2)(28) + (57)(28^2)] = 4.60(28^2) - 57^2(1.35)

33. amistre64

4.60(28^2) - 57^2(1.35) b = ----------------------; thats b lol (57)(28^2) - (57^2)(28)

34. anonymous

wow

35. amistre64

a = 1.60 - .25 -b(28) --------------- ; now plug the value of b into this spot (28)^2

36. amistre64

and thats a

37. amistre64

3606.40 - 4386.15 ----------------- = b 44688 - 90972

38. amistre64

b = 0.016847074583009247256071212514044 if I kept track of it... compare that to the one I posted earlier

39. amistre64

+ 0.011765404891539196; this ones better, I probably missed something earlier....

40. anonymous

ok

41. amistre64

but if you must show your work, its just a matter of grunging your way thru it..

42. amistre64

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

n1 = year1 ; p1 = pay1 and the rest is after this manner

44. anonymous

Lol. Parentheses fun. I always like seeing things solved for variables though.

45. amistre64

$\frac{[(p1-p2)]*[(n3^2)-(n1^2)]+[(p3-p1)]*[(n2^2)-(n1^2)]}{[(n3^2)-(n1^2)]*[(n1-n2)]+[(n2^2)-(n1^2)]*[(n3-n1)]}$

46. amistre64

thats 'b' lol

47. amistre64

you also wanna know when it was near $1 right? 48. amistre64 1959 is under a 1 and 1960 is a little over 1 49. amistre64 any questions yet? ;) 50. amistre64 (p2-p1) (n2-n1)*b a = ------------ - ------------- (n2^2)-(n1^2) (n2^2)-(n1^2) 51. amistre64 (p2-p1) - (n2-n1)*b a = ---------------------- is just as good (n2^2)-(n1^2) 52. anonymous 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 no, thats not correct minwage = y = 1$1 is not a constant for the equation; it is a solution to it

54. amistre64

and you are missing your middle term

55. amistre64

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 so could you show me how to write the equation and i try to solve it again 57. amistre64 or trial and error it like I did ;) 58. amistre64 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 ok 60. amistre64 y = year and p = price; makes sense? a(y1)^2 + b(y1) + c = (p1) 61. amistre64 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 eq1; c = (p1) -b(y1) - a(y1)^2 63. amistre64 laptop cord got unplugged lol 64. amistre64 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 forget that last part, its seems a bit spurious.... forgot to delete it lol 66. amistre64 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 we end up solving this for b and the plugging in the values that I already postd for B above :) 68. amistre64 you feel up to all that? 69. anonymous have to be up to it lol 70. amistre64 I posted all the formulas; start with b go to a and then...well c is .25 for this one lol 71. amistre64 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 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

i might have jacked it up sorry

74. amistre64

If it works for you use it ;) but dont use y for year and y for a dependant variable, its confusing lol

75. amistre64

the way you solve for it depends on the degree of accuracy you are looking for.

76. anonymous

so take y out? do i do them correct

77. anonymous

did i do them correct

78. amistre64

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

ok

80. amistre64

my way gives me 7.259 for min wage in 2007

81. amistre64

compared to 7.25 thats pretty accurate

82. amistre64

1960 gives me 1.006

83. anonymous

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

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

85. amistre64

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

and that aint what you did there :)

87. anonymous

okay lets scratch last equation lol

88. amistre64

do you know how to solve for a system of equations?

89. amistre64

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

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

(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

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

((4.90 * (28^2)) - ((58^2) * 1.35)) / (((-(58^2)) * 28) + (58 * (28^2))) = 0.014363711 Google says b = .014363711

94. amistre64

a = [1.35 - b(28)]/(28^2) (1.35 - (.014363711 * 28)) / (28^2) = 0.0012089491

Find more explanations on OpenStudy