Question

Minimum Wage The table shows the minimum wage for
three different years. Year/wages: 1940/0.25, 1968/1.60, 1997/5.15
(a) Make a scatterplot of the data in the viewing rectangle
[1930, 2010, 10] by [0, 6, 1].
(b) Find a quadratic function given by
that models the data.
f(x)=a(x - h)^2 + k
(c) Estimate the minimum wage in 1976 and compare
it to the actual value of $2.30.
(d) Estimate when the minimum wage was $1.00.
(e) If current trends continue, predict the minimum
wage in 2009. Compare it to the projected value of
$7.25.

Answer 1

can someone help me solve this so i can understand it

Answer 2

it looks like they give you 3 points of reference, right?

Answer 3

yea

Answer 4

this turn into a system of 3 equations with 3 unknown variables; the unknowns are the constant coefficients of our quadratic equation that we are trying to find

Answer 5

lets adjust the years so that the computations are smaller, but equivalent

1940/0.25, 1968/1.60, 1997/5.15
-1940 -1940 -1940
-------------------------------
0, 0.25 28, 1.60 57, 5.15

in essense, all ive done by this is to shift the graph so to the left; same shape, just a different location

Answer 6

makes sense so far?

Answer 7

yes so far

Answer 8

i could also move the graph up or down if need be by adjusting the "y" values, but they look simple enough that i wont try that .... yet :)

Answer 9

oh boy ok math isn't my strongest subject

Answer 10

0, 0.25 28, 1.60 57, 5.15

when x=0, y = 0.25 ; this is our y intercept by definition, whenever x = 0 we have a y intercept

ax^2 +bx + 0.25 = 0.25 ; when x = 0

ax^2 +bx + .25 = 1.60 ; when x = 28

ax^2 +bx + .25 = 5.15 ; when x = 57

Answer 11

if math aint your strongest subject then i take it trying a matrix on this wont make alot of sense to you :)

Answer 12

probably not..i can usally figure it out but for some reason i can't wrap my head around this problem i appreciate your help

Answer 13

lets apply our x values in each case and see if we can solve for a and b

0a +0b + 0.25 = 0.25 ; this ones not going to give us any new information so we can ignore it for now

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

28^2a +28b = 1.35
57^2a +57b = 4.90

now we are left with a system of 2 equations in 2 unknowns to solve for a and b

Answer 14

if we forget about trying to find the smallest value of this and that, we can simply eliminate some prospects by multiplying the top by 57^2 and the bottom by -28^2


28^2 (57^2)a +28 (57^2)b = 1.35(57^2)
57^2(-28^2)a +57(-28^2)b = 4.90(-28^2)
---------------------------------------
0 +(28 (57^2)+57(-28^2))b = 1.35(57^2) + 4.90(-28^2)

dividing off that messy stuff gives us:

\[b=\frac{1.35(57^2) + 4.90(-28^2)}{28 (57^2)+57(-28^2) }\]

Answer 15

if the problem is ax^2+bx +.25 would the equation be a(28)^2+b(28)+.25? or am i wrong in assuming that?

Answer 16

with the aid of a calculator; b = abt .012

and the same concept can be applied to get rid of the b to solve for a

Answer 17

your right, I just subtracted the .25 from each side to get a and b all alone

Answer 18

ok im confused

28^2 (57^2)a +28 (57^2)b = 1.35(57^2)
57^2(-28^2)a +57(-28^2)b = 4.90(-28^2)
---------------------------------------
0 +(28 (57^2)+57(-28^2))b = 1.35(57^2) + 4.90(-28^2)

dividing off that messy stuff gives us:

\[b=\frac{1.35(57^2) + 4.90(-28^2)}{28 (57^2)+57(-28^2) }\]

this comes from ?

Answer 19

28^2(57)a + 28(57)b = 1.35(57)
57^2(-28)a +57(-28)b = 4.90(-28)
-------------------------------
(28^2(57)+57^2(-28))a +0 = 1.35(57) + 4.90(-28)

\[a=\frac{1.35(57) + 4.90(-28)}{28^2(57)+57^2(-28)}=-60.25\]

ill explain the best i can after i post this bit of it

Answer 20

the numbers are messy, but the concept is still to use elimination to get rid of one of the variables in order to solve whats left for the remaing variable

Answer 21

\[\begin{array}\ a_1x+b_1c=n\\a_2x+b_2c=m\end{array}\]

\[\begin{array}\ (a_1x+b_1c=n)*a_2\\(a_2x+b_2c=m)*(-a_1)\end{array}\]

\[\begin{array}\ \cancel{a_1a_2}x+a_2b_1c=a_2n\\\bcancel{-a_1a_2}x-a_1b_2c=-a_1m\\------------\\(a_2-a_1)b_2c=a_2n-a_1m\end{array}\]

Answer 22

ugh, forgot a few parts in that; but it should still make sense i hope lol

Answer 23

not really im an epic fail at this point of no sleep

Answer 24

thats ok, these things can get complicated; i just tested out my "solution" and apparently i missed it someplace.

Answer 25

ok well im lost so i cant help you much :/

Answer 26

i hit a wrong button for the "a"; with any luck its about .001

so i want to test out the following to see how close I get to the points given

y = .001x^2 +.012x+.25

Answer 27

ok i'll be right here :)

Answer 28

its yelling at me :/ with my approximations the wolf is giving me a value that is .25 cents less.

ima gonna try the matrix route and see what it gives me ....

but the concepts i presented are sound, its just working the numbers themselves thats a bit helter skelter

Answer 29

oh don't let it get you like it has gotten me I still need to explain this to others I am working with so the better you are the more it helps me :) cuz right now i am a deer in head lights just looking at the equation coming

Answer 30

http://www.wolframalpha.com/input/?i=rref%7B%7B0%2C0%2C1%2C.25%7D%2C%7B28%5E2%2C28%2C1%2C1.60%7D%2C%7B57%5E2%2C57%2C1%2C5.15%7D%7D

well, this is what it pops out to

Answer 31

which is what i had to begin with .... so i think the wolf is getting a little mixed up these days

Answer 32

oh my

Answer 33

:/

Answer 34

the "exact" values for our coeefs are monster looking fractions; and my approximation to those is making my quadratic miss the mark the further it moves away from 1940

Answer 35

:( oh no

Answer 36

\[y = \frac{241}{185136}(x-1940)^2 +\frac{10891}{925680}(x-1940)+\frac{1}{4}\]

that works lol

Answer 37

omg what is that

Answer 38

thats the quadratic equation that fits your data points; i got no idea how that would look in the a(x-h)^2+k format tho

Answer 39

hmm

Answer 40

for x=1997 i get 2.36; and the actual they presented in 2.30

Answer 41

it estimates about 1960 for a minimum wage of 1.00

Answer 42

and it estimates 7.26 for 2009

Answer 43

ok well thats awesome thank you know i just have to figure out how I am going to show and explain it

Answer 44

i think youd have to use computer generated values in order to "show" it; but the explanation is the basic notion that i started with

Answer 45

we create a system of 3 equations in 3 unknown and solve it using approved methods

Answer 46

either substitution, elimination, or matrix methods

Answer 47

ok thank you at least you got me headed in the right direction :)

Answer 48

lol, good luck with it ;)

Answer 49

thanks