anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
ack!! not this one again..I got a program for this ;)
anonymous
  • anonymous
lol
amistre64
  • amistre64
i programmed this just for cases like this
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
can you help me solve for a
amistre64
  • amistre64
yep
anonymous
  • anonymous
@amistre64 :D
amistre64
  • amistre64
0.001301745743669519x^2 -5.0390080805461945x + 4876.675395385017
amistre64
  • amistre64
:)
amistre64
  • amistre64
a = something like 64.24/48957 or some rediculaous monster of a fraction
amistre64
  • amistre64
you wanna know the min wage of 1976 right?
anonymous
  • anonymous
so when i re write the equation it will look like f(x)= a(x-1940)^2+0.25
anonymous
  • anonymous
yes
amistre64
  • amistre64
screenshot
1 Attachment
amistre64
  • amistre64
1976 = 2.361
amistre64
  • amistre64
you can write the equation to be a(x-40)^2 to make the numbers easier yes
anonymous
  • anonymous
so which of the numbers would i put in for a being that i have to show steps
amistre64
  • amistre64
if we adjust in my program for the -40 we get this: 0.0013017457436695185(x-40)^2 + 0.011765404891539196(x-40) + 0.25
amistre64
  • amistre64
ive got a post about this same thing a couple of days ago that I typed it all out :)
anonymous
  • anonymous
thanks
anonymous
  • 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.
amistre64
  • amistre64
id have to redo it all, its a monster tho....here goes :)
amistre64
  • amistre64
a(1940-1940)^2 + b(1940-1940) + c = .25 from this we see that c = .25 right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
im gonna write it as the lesser numbers instead of the years ok? just be easier...
anonymous
  • anonymous
ok
amistre64
  • amistre64
the other 2 equations then are: a(28)^2 +b(28) + .25 = 1.60 a(57)^2 +b(57) + .25 = 5.15
amistre64
  • amistre64
solve for a I suppose is as likely as any other...
amistre64
  • amistre64
a = 1.60 - .25 -b(28) --------------- now use this value in the other one (28)^2
amistre64
  • amistre64
write down that value for a, your gonna need it :)
amistre64
  • amistre64
a(57^2) + b(57) = 5.15 - .25 (57^2)(1.60 - .25 -b(28) ----------------------- + b(57) = 4.60 (28)^2
amistre64
  • amistre64
(57^2)(1.35) -b(57^2)(28) + b(57)(28^2) = 4.60(28^2)
amistre64
  • amistre64
b[-(57^2)(28) + (57)(28^2)] = 4.60(28^2) - 57^2(1.35)
amistre64
  • amistre64
4.60(28^2) - 57^2(1.35) b = ----------------------; thats b lol (57)(28^2) - (57^2)(28)
anonymous
  • anonymous
wow
amistre64
  • amistre64
a = 1.60 - .25 -b(28) --------------- ; now plug the value of b into this spot (28)^2
amistre64
  • amistre64
and thats a
amistre64
  • amistre64
3606.40 - 4386.15 ----------------- = b 44688 - 90972
amistre64
  • amistre64
b = 0.016847074583009247256071212514044 if I kept track of it... compare that to the one I posted earlier
amistre64
  • amistre64
+ 0.011765404891539196; this ones better, I probably missed something earlier....
anonymous
  • anonymous
ok
amistre64
  • amistre64
but if you must show your work, its just a matter of grunging your way thru it..
amistre64
  • 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))
amistre64
  • amistre64
n1 = year1 ; p1 = pay1 and the rest is after this manner
anonymous
  • anonymous
Lol. Parentheses fun. I always like seeing things solved for variables though.
amistre64
  • amistre64
\[\frac{[(p1-p2)]*[(n3^2)-(n1^2)]+[(p3-p1)]*[(n2^2)-(n1^2)]}{[(n3^2)-(n1^2)]*[(n1-n2)]+[(n2^2)-(n1^2)]*[(n3-n1)]}\]
amistre64
  • amistre64
thats 'b' lol
amistre64
  • amistre64
you also wanna know when it was near $1 right?
amistre64
  • amistre64
1959 is under a 1 and 1960 is a little over 1
amistre64
  • amistre64
any questions yet? ;)
amistre64
  • amistre64
(p2-p1) (n2-n1)*b a = ------------ - ------------- (n2^2)-(n1^2) (n2^2)-(n1^2)
amistre64
  • amistre64
(p2-p1) - (n2-n1)*b a = ---------------------- is just as good (n2^2)-(n1^2)
anonymous
  • 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
amistre64
  • amistre64
no, thats not correct minwage = y = 1 $1 is not a constant for the equation; it is a solution to it
amistre64
  • amistre64
and you are missing your middle term
amistre64
  • 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
anonymous
  • anonymous
so could you show me how to write the equation and i try to solve it again
amistre64
  • amistre64
or trial and error it like I did ;)
amistre64
  • 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 :)
anonymous
  • anonymous
ok
amistre64
  • amistre64
y = year and p = price; makes sense? a(y1)^2 + b(y1) + c = (p1)
amistre64
  • 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
amistre64
  • amistre64
eq1; c = (p1) -b(y1) - a(y1)^2
amistre64
  • amistre64
laptop cord got unplugged lol
amistre64
  • 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
amistre64
  • amistre64
forget that last part, its seems a bit spurious.... forgot to delete it lol
amistre64
  • 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]
amistre64
  • amistre64
we end up solving this for b and the plugging in the values that I already postd for B above :)
amistre64
  • amistre64
you feel up to all that?
anonymous
  • anonymous
have to be up to it lol
amistre64
  • amistre64
I posted all the formulas; start with b go to a and then...well c is .25 for this one lol
amistre64
  • 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)]
anonymous
  • 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
anonymous
  • anonymous
i might have jacked it up sorry
amistre64
  • amistre64
If it works for you use it ;) but dont use y for year and y for a dependant variable, its confusing lol
amistre64
  • amistre64
the way you solve for it depends on the degree of accuracy you are looking for.
anonymous
  • anonymous
so take y out? do i do them correct
anonymous
  • anonymous
did i do them correct
amistre64
  • amistre64
I dont know how you did them ;) but you shouldnt use y to mean year AND y to mean value of the function....
anonymous
  • anonymous
ok
amistre64
  • amistre64
my way gives me 7.259 for min wage in 2007
amistre64
  • amistre64
compared to 7.25 thats pretty accurate
amistre64
  • amistre64
1960 gives me 1.006
anonymous
  • 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
amistre64
  • amistre64
i dont understand the process of how you get your equation; im sure it has to do with homogeneous difference equattions tho...
amistre64
  • 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
amistre64
  • amistre64
and that aint what you did there :)
anonymous
  • anonymous
okay lets scratch last equation lol
amistre64
  • amistre64
do you know how to solve for a system of equations?
amistre64
  • 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....
amistre64
  • 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
amistre64
  • 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)
amistre64
  • 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'
amistre64
  • amistre64
((4.90 * (28^2)) - ((58^2) * 1.35)) / (((-(58^2)) * 28) + (58 * (28^2))) = 0.014363711 Google says b = .014363711
amistre64
  • amistre64
a = [1.35 - b(28)]/(28^2) (1.35 - (.014363711 * 28)) / (28^2) = 0.0012089491

Looking for something else?

Not the answer you are looking for? Search for more explanations.