A math problem.

[Green_Tea]

I am working on a function. The domain and range values are as follows:

{(1,4), (2,12), (3,24), (4,40)}

I need to find the equation that would make it possible to solve for the nth term.

I have no idea how to go about finding the equation - I believe that I can use a graphing calculator to do so, but do not know how to use one. I also think I can somehow use the quadratic equation, but again, no idea how.

HELP.

(This is for a math class I am taking to get a Master's in elementary ed. I have a test tomorrow, and this is the only concept I cannot figure out.)

[nrp]

I messed around with it and came up with x(2x +2), which I think works. It was trial and error, though, so I don't know how to tell you to replicate it, if that's what you need to do.

[Green_Tea]

Thanks, nrp. That's super helpful for this problem, but unfortunately I do need to be able to apply the concept to other problems.

This math class is sucking the life out of me :(.

[larig]

X Y
(1 , 4)
...........>8
(2, 12)..........>(12-8) =4
............>12
(3, 24)...........>(16-12) =4
...........>16
(4, 40)



Okay so my work is above. I just rearranged the data into table form. Then I found the "successive differences" between the y-values (because we have Xs that step by 1). (e.g., the difference between 12-4 = 8). Then I find the differences between the differences I just found. I keep going until I wind up with the same difference, in this case 4. Since it took me two successive differences I have QUADRATIC data--that would be fit by an equation of the form y = ax^2 + bx + c. If the first differences had been the same you'd have had linear related data points/ordered pairs.

What kind of graphing calculator do you have? (I'll tell you how to find the equation based on your answer to this--I'll also tell you how to do by hand, wanted to give you a chance to digest this first.

Does the above make sense?

[Green_Tea]

X Y
(1 , 4)
...........>8
(2, 12)..........>(12-8) =4
............>12
(3, 24)...........>(16-12) =4
...........>16
(4, 40)



Okay so my work is above. I just rearranged the data into table form. Then I found the "successive differences" between the y-values (because we have Xs that step by 1). (e.g., the difference between 12-4 = 8). Then I find the differences between the differences I just found. I keep going until I wind up with the same difference, in this case 4. Since it took me two successive differences I have QUADRATIC data--that would be fit by an equation of the form y = ax^2 + bx + c. If the first differences had been the same you'd have had linear related data points/ordered pairs.

What kind of graphing calculator do you have? (I'll tell you how to find the equation based on your answer to this--I'll also tell you how to do by hand, wanted to give you a chance to digest this first.

Does the above make sense?

YES, it makes sense!! In fact, I got that far. I just don't understand how to use the quadratic equation to reach the equation to find the nth term.

I don't actually own a graphing calc. I could conceivably download one for my ipod touch, but I am not sure how steep the learning curve is for using one. I have a test tomorrow, so perhaps learning how to do it by hand is my best bet. But I am 100% willing to download the calc if you think it's the easiest and most accurate way to get the answer.

THANK YOU!!!

[KrisM]

I'd solve it like this:

ax^2 + bx + c = y

substituting the numbers for x and y from your sets:
a(1)^2 + b(1) + c = 4
a(2)^2 + b(2) + c = 12
etc.

which become
a + b + c = 4
4a + 2b + c = 12
9a + 3b + c = 24
16a + 4b + c =40

I would take the first 2 equations and subtract the first from the second. I'd then subtract the 4th from the 3rd. This would give me 2 new equations:
3a + b = 8
7a + b = 16
Subtract the 1st from the 2nd to get:
4a = 8
a = 2
plug the a=2 into one of the last 2 equations to get b = 2. Put a=b=2 into the very first equation to find c=0.
ax^2 + bx + c = y
2x^2 + 2x = y

[larig]

by hand is fine. I'll show you a pretty straight forward way to do it.

We need to figure out what A, B and C are in this equation: y = Ax^2 + Bx + c

You use the points you were given. (just the easiest 3 to plug in).

(1, 4) --> 4 = A (1)^2 + B(1) + C--->(simplify) --> 4 = A + B + C (eq. 1)

(2, 12) -->12 = A (2)^2 + B (2) + C --> (simplify) --> 12 = 4A + 2B + C (eq. 2)

(3, 24) --> 24 = A (3)^2 + B (3) + C --> (simplify) --> 24 = 9A + 3B + C (eq. 3)

with me?

[Green_Tea]

by hand is fine. I'll show you a pretty straight forward way to do it.

We need to figure out what A, B and C are in this equation: y = Ax^2 + Bx + c

You use the points you were given. (just the easiest 3 to plug in).

(1, 4) --> 4 = A (1)^2 + B(1) + C--->(simplify) --> 4 = A + B + C (eq. 1)

(2, 12) -->12 = A (2)^2 + B (2) + C --> (simplify) --> 12 = 4A + 2B + C (eq. 2)

(3, 24) --> 24 = A (3)^2 + B (3) + C --> (simplify) --> 24 = 9A + 3B + C (eq. 3)

with me?

Yup, got it...

[mom2binsd]

Holy cow, I have no idea what you all are talking about!

Math is NOT my thing. Impressed by all you math skills, especially so late at night!

[Green_Tea]

OMG, I think I got it. I figured it out AND it matches the answer in the back of the book. You guys are awesome. Thank you SO MUCH!

[larig]

PP's work looks good, I'll just elaborate some more and present some alternative methods for solving.

What you have is a system of equations in three variables (A, B, & C). There are lots of ways to solve these.

one way is just plain old algebraically (like PP did).

take eq 1 and isolate C. C = 4 - A - B.
then substitute this in for c in eqs. 2 & 3.

(2) subst. for C 12 = 4A + 2B + (4 - A - B)
simplify (combine like terms) 12 = 3A + B + 4
simplify (subt. 4 from both sides) 8 = 3A + B


do same for equation 3.
(3) subst. for C 24 = 9A + 3B + (4 - A - B)
simplify (combine like terms) 24 = 8A + 2B + 4
simplify (subt. 4 from both sides) 20 = 8A + 2B

Then I take the new equations 2 & 3 and add them together. I'm going to try to eliminate the Bs, so to do that I need to have a -2B in the second equation. So, I'll multiply equation 2 by -2]

eq 2: 8(-2) = 3A(-2) + B (-2)
eq 3: 20 = 8A + 2B
----------------------------
-16 = -6A - 2B
20 = 8A + 2B
-----------------------Add the two equations
4 = 2A
----------------now solve for A by dividing by 2
2 = A

with me?

ETA: I stopped because OP said she got it.

[larig]

OMG, I think I got it. I figured it out AND it matches the answer in the back of the book. You guys are awesome. Thank you SO MUCH!

:jammin:

I am so happy to help if you ever need it again.

[Green_Tea]

:jammin:

I am so happy to help if you ever need it again.

Careful what you wish for ;).

In all seriousness, you have no idea how much that means to me. This class has been a soul sucking experience, not because I don't think I am capable of grasping the material but because the professor is terrible. Having someone be able to explain it to me in an accessible way feels like the sun breaking through the gloom.

Thank you!!

[egoldber]

Have you heard of this site?

http://www.khanacademy.org

It's a collection of YouTube videos on virtually any math topic you can think of. I heard the founder on NPR one day and looked at it. It's truly an incredible endeavor!

http://www.khanacademy.org/about

[boogiemomz]

Holy cow, I used to be really good at math and I don't know how to do any of that. Quadratic equation and graphing calculators in elementary ed???? Yikes!!!:dizzy: So, you decided to stick out the class? I hope you get through it in a way that is satisfactory for you!! Sounds miserable!!! :hug:

[crayonblue]

This brings back great memories for me. I LOVED math!!!!!!!!!! Now the extent of my math is figuring out cost per diaper. :)

[brittone2]

Have you heard of this site?

http://www.khanacademy.org

It's a collection of YouTube videos on virtually any math topic you can think of. I heard the founder on NPR one day and looked at it. It's truly an incredible endeavor!

http://www.khanacademy.org/about

TFS! As a HSer I love that there are more sites like this out there now...will be great as the kids get older. It has been a while since I've had to do any challenging math and these will be great as refreshers down the road.

[egoldber]

And so far he has done all the videos himself, which I think is pretty astounding!

[brittone2]

And so far he has done all the videos himself, which I think is pretty astounding!
Amazing!

[elizabethkott]

My head just imploded reading this thread.
There's a reason I teach theatre. :D

[hillview]

WOW. I need more coffee for this thread. You all rock!
/hillary

[Green_Tea]

Have you heard of this site?

http://www.khanacademy.org



I'll check it out - thanks!


Holy cow, I used to be really good at math and I don't know how to do any of that. Quadratic equation and graphing calculators in elementary ed???? Yikes!!!:dizzy: So, you decided to stick out the class? I hope you get through it in a way that is satisfactory for you!! Sounds miserable!!! :hug:

I decided to stick it out because on the last day to drop I got back a test and my grade was better than I had anticipated. I am still hovering around a 75 (which to me is very disappointing because I literally have A+s in my 3 other classes) but I am hoping to eek out an 80 for a B- on my transcript. While I think the work is challenging, the real problem lies with the professor, who insists that the reason that the majority of the class is failing (I have the second or third highest grade in the entire class) is because we're just not working hard enough. I spend more time on this class than the other three combined - I couldn't work harder if I tried! A lot of what we're learning is brand new to us - performing functions in bases other than base 10, working with the Mayan, Egyptian and Babylonian number systems, etc. - yet he acts as though we're all stupid if takes some review to understand. Luckily I am taking the followup class - Math 106 - (next semester) with a MUCH better teacher. In the meantime, please send all your good math vibes this way!!


My head just imploded reading this thread.
There's a reason I teach theatre. :D

I am green with envy. My undergrad degree is in theatre :love-retry:. If I thought I would have a snowball's chance in hell of finding a full time teaching job with a theatre cert, I'd totally be working on that!

Thank you all for your support. It means a lot!!

[larig]

Careful what you wish for ;).

In all seriousness, you have no idea how much that means to me. This class has been a soul sucking experience, not because I don't think I am capable of grasping the material but because the professor is terrible. Having someone be able to explain it to me in an accessible way feels like the sun breaking through the gloom.

Thank you!!

Green Tea, I taught HS math for 10 years before starting work on my PhD. I sometimes really miss it. Helping you tonight was good for my math-loving soul. I am truly serious.

I would love to help you though the rest of this semester--glad you stuck it out. You can totally do this!

ETA: what textbook are you using?

[Green_Tea]

Green Tea, I taught HS math for 10 years before starting work on my PhD. I sometimes really miss it. Helping you tonight was good for my math-loving soul. I am truly serious.

I would love to help you though the rest of this semester--glad you stuck it out. You can totally do this!

ETA: what textbook are you using?

The text is called A Problem Solving Approach to Mathematics for Elementary School Teachers. It's by Billstein, Libeskind and Lott.

It's a decent book, but has some huge gaps.

Again, thank you. I just got home from my test and I think I actually did OK!

[larig]

Great news about the test!

I just wondered if the text was one I'd recognize--I don't (that's neither here nor there!)

[Corie]

My head just imploded reading this thread.
There's a reason I teach theatre. :D

I was just thinking the exact same thing.
Except I used to teach Home Ec.
The only math I need is doubling a recipe. :)