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.

- anonymous

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- anonymous

can someone help me solve this so i can understand it

- amistre64

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

- anonymous

yea

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## More answers

- amistre64

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

- amistre64

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
|dw:1332971979117:dw|

- amistre64

makes sense so far?

- anonymous

yes so far

- amistre64

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 :)

- anonymous

oh boy ok math isn't my strongest subject

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

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?

- amistre64

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

- amistre64

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

- anonymous

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 ?

- amistre64

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

- amistre64

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

- amistre64

\[\begin{array}\ a_1x+b_1c=n\\a_2x+b_2c=m\end{array}\]
\[\begin{array}\ (a_1x+b_1c=n)*a_2\\(a_2x+b_2c=m)*(-a_1)\end{array}\]
\[\begin{array}\ \cancel{a_1a_2}x+a_2b_1c=a_2n\\\bcancel{-a_1a_2}x-a_1b_2c=-a_1m\\------------\\(a_2-a_1)b_2c=a_2n-a_1m\end{array}\]

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

ok i'll be right here :)

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

oh my

- anonymous

:/

- amistre64

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

- anonymous

:( oh no

- amistre64

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

- anonymous

omg what is that

- amistre64

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

- anonymous

hmm

- amistre64

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

- amistre64

it estimates about 1960 for a minimum wage of 1.00

- amistre64

and it estimates 7.26 for 2009

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

either substitution, elimination, or matrix methods

- anonymous

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

- amistre64

lol, good luck with it ;)

- anonymous

thanks

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