Question

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

Answer 1

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

Answer 2

lol

Answer 3

i programmed this just for cases like this

Answer 4

can you help me solve for a

Answer 5

yep

Answer 6

@amistre64 :D

Answer 7

0.001301745743669519x^2
-5.0390080805461945x
+ 4876.675395385017

Answer 8

:)

Answer 9

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

Answer 10

you wanna know the min wage of 1976 right?

Answer 11

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

Answer 12

yes

Answer 13

screenshot

Answer 14

1976 = 2.361

Answer 15

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

Answer 16

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

Answer 17

if we adjust in my program for the -40 we get this:

0.0013017457436695185(x-40)^2
+ 0.011765404891539196(x-40)
+ 0.25

Answer 18

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

Answer 19

thanks

Answer 20

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.

Answer 21

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

Answer 22

a(1940-1940)^2 + b(1940-1940) + c = .25

from this we see that c = .25 right?

Answer 23

yes

Answer 24

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

Answer 25

ok

Answer 26

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

a(57)^2 +b(57) + .25 = 5.15

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

wow

Answer 35

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

Answer 36

and thats a

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

ok

Answer 41

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

Answer 42

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))

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

thats 'b' lol

Answer 47

you also wanna know when it was near $1 right?

Answer 48

1959 is under a 1 and 1960 is a little over 1

Answer 49

any questions yet? ;)

Answer 50

(p2-p1) (n2-n1)*b
a = ------------ - -------------
(n2^2)-(n1^2) (n2^2)-(n1^2)

Answer 51

(p2-p1) - (n2-n1)*b
a = ---------------------- is just as good
(n2^2)-(n1^2)

Answer 52

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

Answer 53

no, thats not correct minwage = y = 1

$1 is not a constant for the equation; it is a solution to it

Answer 54

and you are missing your middle term

Answer 55

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

Answer 56

so could you show me how to write the equation and i try to solve it again

Answer 57

or trial and error it like I did ;)

Answer 58

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 :)

Answer 59

ok

Answer 60

y = year and p = price; makes sense?

a(y1)^2 + b(y1) + c = (p1)

Answer 61

a(y1)^2 + b(y1) + c = (p1) Eq1
a(y2)^2 + b(y2) + c = (p2) Eq2
a(y3)^2 + b(y3) + c = (p3) Eq3

Answer 62

eq1; c = (p1) -b(y1) - a(y1)^2

Answer 63

laptop cord got unplugged lol

Answer 64

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

Answer 65

forget that last part, its seems a bit spurious.... forgot to delete it lol

Answer 66

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]

Answer 67

we end up solving this for b and the plugging in the values that I already postd for B above :)

Answer 68

you feel up to all that?

Answer 69

have to be up to it lol

Answer 70

I posted all the formulas; start with b go to a and then...well c is .25 for this one lol

Answer 71

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)]

Answer 72

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

Answer 73

i might have jacked it up sorry

Answer 74

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

Answer 75

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

Answer 76

so take y out? do i do them correct

Answer 77

did i do them correct

Answer 78

I dont know how you did them ;) but you shouldnt use y to mean year AND y to mean value of the function....

Answer 79

ok

Answer 80

my way gives me 7.259 for min wage in 2007

Answer 81

compared to 7.25 thats pretty accurate

Answer 82

1960 gives me 1.006

Answer 83

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

Answer 84

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

Answer 85

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

Answer 86

and that aint what you did there :)

Answer 87

okay lets scratch last equation lol

Answer 88

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

Answer 89

c= .25
a(28^2) + b(28) + c = 1.60
a(58^2) + b(58) + c = 5.15

solve by substitution, elimination or matrix....

Answer 90

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

Answer 91

(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)

Answer 92

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'

Answer 93

((4.90 * (28^2)) - ((58^2) * 1.35)) / (((-(58^2)) * 28) + (58 * (28^2))) = 0.014363711

Google says b = .014363711

Answer 94

a = [1.35 - b(28)]/(28^2)


(1.35 - (.014363711 * 28)) / (28^2) = 0.0012089491