Math help

[Kirby]

doesn't excel have a function finder thingy where you put in points on a graph, put the degree of the equation and it'll fit it for you?

 

[Accipiter22]

Originally posted by: nkgreen
doesn't excel have a function finder thingy where you put in points on a graph, put the degree of the equation and it'll fit it for you?

no.






UGH if no one on ATOT knows how to do this I may just have to try randomly guessing at formulas till I get one right. I had a similar issue last week, using different numbers and just kept writing down random formulas till I found one that worked. For all these engineers on here you guys aren't that good at math lol. I guess we're all in the same boat there

 

[Accipiter22]

Originally posted by: nycxandy

Originally posted by: DrPizza
Hopefully, someone reads this post before they simply post the answer for you, like they did last time.

I clearly stated how to solve the problem in both of your other threads. You asked for "help", not "someone do my homework for me and I'll copy down the answer."

Write down ax^2 + bx + c = y three times.
in each equation, replace x and y with one of your data points.

i.e. a*5^2 + b*5 + c = 2
(for the point (5,2) )
This becomes 25a +5b + c = 2

You'll end up with three equations and 3 unknowns. Solve for a, b, and c, and you'll have a quadratic equation that passes exactly through each of those points.

Yeah, this is the perfect way to do it with algebra. Simple and easy.

ok, let's pretend that would even help....so I have 3 formulas with 3 variables each! so how does this help me??? There 3 completely seperate formulas with a bunch of variables in them. How do I solve for them if there's 3 different formulas now??

 

[Accipiter22]

see! you can't solve it with 3 random formulas, ther'es no way to combine them

 

[Soccer55]

Originally posted by: Accipiter22
see! you can't solve it with 3 random formulas, ther'es no way to combine them

You can use substitution to solve for the constants a,b,c. Or if you're feeling zesty, you can use a matrix and gaussian elimination to figure out what a,b,c are. I'm not going to spend the time working it all out, but it should be possible using the method that DrPizza described.

-Tom

 

[Accipiter22]

I don't know HOW to use substitution!!! I don't know what that is, what it means, I have never heard of matrix or guassian elimination. if I KNEW any of these things, I wouldn't need ATOT's help I imagine.

 

[Howard]

I learned substitution in grade 10...

 

[Accipiter22]

I learned you're a douchebag 2 minutes ago

 

[jagec]

Originally posted by: Accipiter22

Originally posted by: nkgreen
doesn't excel have a function finder thingy where you put in points on a graph, put the degree of the equation and it'll fit it for you?

no.

UGH if no one on ATOT knows how to do this I may just have to try randomly guessing at formulas till I get one right. I had a similar issue last week, using different numbers and just kept writing down random formulas till I found one that worked. For all these engineers on here you guys aren't that good at math lol. I guess we're all in the same boat there

Excel DOES fit points. Proof.

Believe me, none of us are UNABLE to do your problem. We're trying to walk you through step-by-step, but you have to do SOME of the work too. Insults get you nowhere.

Originally posted by: Accipiter22
I learned you're a douchebag 2 minutes ago

Eh, forget it. Learn some humility, or learn to do your own homework.

 

[Howard]

Originally posted by: Accipiter22
I learned you're a douchebag 2 minutes ago

waaah waah waahhh

Go find a tutor.

 

[Elderly Newt]

Originally posted by: Accipiter22

Originally posted by: DrPizza
Hopefully, someone reads this post before they simply post the answer for you, like they did last time.

I clearly stated how to solve the problem in both of your other threads. You asked for "help", not "someone do my homework for me and I'll copy down the answer."

Write down ax^2 + bx + c = y three times.
in each equation, replace x and y with one of your data points.

i.e. a*5^2 + b*5 + c = 2
(for the point (5,2) )
This becomes 25a +5b + c = 2

You'll end up with three equations and 3 unknowns. Solve for a, b, and c, and you'll have a quadratic equation that passes exactly through each of those points.

that may be easy for someone who holds a doctorate in Pizza....but I have NO idea how to solve for that stuff that you gave me......and again, you're reversing the equation...I'm NOT going to have the (0,1,2)...I'll be given a number between 0.0 and 7.0. so having an equation that =2 doesn't do me any good. I can't solve that equation you gave me since there's an x^2 and a regular X in it and I don't know how to do that.


To answer the other questions, the reward can be anywhere between 0 and upward, and with as many decimals as the formula gives.

Ok... so you have 3 x values and 3 corresponding y values.
Dr.Pizza is saying you need to plug in those 3 ordered pairs into the equation ax^2 + bx + c = y

you dont know what a, b, or c is. They are constants; they don't change, but you dont know what they are yet. All you know is x and y. So, like he said, once you plug in one ordered pair, you're going to get something like 25a +5b + c = 2. Save this equation. Now do it 2 more times with the other ordered pairs, and save those other 2 equations.

You now have 3 equations, with 3 unknown variables (a, b, and c). Solve one equation for a variable (c, for example). You will have c = something. Look at your other 2 equations. See those c's? They used to be unknowns, but guess what? Now you know what c is! You just solved for it! So substitute whatever c equals into one of those other equations.

Guess what now? Your c is gone! Now you only have 2 unknowns (a and b). Solve for one of those, just like you did for c. Now, lets say you solve for b. You're going to have b = something. Substitute that for b just like you did back when you solved for c. Now you have an equation with just a, and i think you know how to solve for a.

Work with it. If you don't know what a, b or c is, see if you can solve it in terms of another variable, and go from there. I hope this helped and didnt confuse you...


ok, quick rundown on substitution.
lets say you have these
x + y + z = 32
x = 2
y = 3x
z = 2xy

x=2 is simple enough, no problems
y = 3x isnt so simple, but you know what x equals so
y = 3(2) = 6
same deal with z
z = 2(2)(6) = 24
go back to the original equation and substitute
2 + 6 + 24 = 32

you need to do the same thing in your problem, except its more complicated. same basic idea though...

and also... its very very strange that you're doing this without having been taught what substitution is...

 

[mackle]

How about if A=even #(0?,2,4,6,...ect)....B=Ax2.5...and if A=odd # (1,3,5,7...ect) B=Ax3 ...??Can't think of the formula

 

[Screech]

Using a calculator with quadreg will spit out the answer in about 2 seconds. Perhaps I will bless you with the answer if you stop calling people douchebags.......

 

[dullard]

Originally posted by: Accipiter22

Originally posted by: nkgreen
doesn't excel have a function finder thingy where you put in points on a graph, put the degree of the equation and it'll fit it for you?

no.

Accipiter22, Excel DOES have this function as jagec pointed out. And it is very easy to use. I strongly urge you to learn to use it. Excel is far more powerful than most people give it credit for.

 

[dullard]

Since Accipiter22 seems unable to try the confusing directions above, I'll actually show them to him. There is no need to beat around the bush like everyone else here is doing. I'll use DrPizza's formula and do it step by step.

Accipiter22, I really hope you read this and try it yourself. I took the time to type it out, I hope you take the time to write it out yourself with a piece of paper and solve it with me.

Assumptions:
[*]Assume you want a polynomial.
[*]Assume you want a polynomial that exactly matches all the points you gave.

Then, with those assumptions, we can continue:
[*]Since you have 3 known points, that polynomial MUST have 3 terms to be guaranteed to fit any 3 points. You may be lucky with fewer terms, but in most cases, you need 3 terms for 3 points. If you had 5 points, you'd need 5 terms.
[*]DrPizza gave a clear example of a polynomial with three terms. However, he made it very confusing. So let me make it less confusing. Lets use this polynomial:

• B = c + d*A + e*A*A.

[*]Do you see how there are three terms? "c" is one term. "d*A" is another term. "e*A*A" is the third term. If you wanted more terms for more known points, just continue the pattern.

See if you have enough data to solve the problem:
[*]That equation above has three missing pieces of information. What do "c", "d", and "e" equal? We just don't know yet.
[*]You have three independent pieces of information: B=0 when A=2.5, B=1 when A=4, and B=2 when A=5.
[*]Since you have three data points and three unknowns, this can be solved easilly. IF these numbers don't match, it is much more difficult to solve. But since they do match, this is a ~8th grade math problem (give or take a year depending on your school).

Plug in the data into the equation:
[*]We will do this three times because we have three unknowns and three data points.
[*]#1: A=2.5, B=0. Thus, plugging it into the equation:

• 0 = c + d*2.5+ e*2.5*2.5.

[*]#2: A=4, B=1. Thus, plugging it into the equation:

• 1 = c + d*4+ e*4*4.

[*]#3: A=5, B=2. Thus, plugging it into the equation:

• 2 = c + d*5+ e*5*5.

**Solve the equations for "c", "d", or "e":
[*]There are many, many ways to proceed, I will show you one possible way to solve this. Other ways may be easier or harder for you. But all you'll ever need to know in life is one method. Learn it, and you'll be able to solve everything. Math teachers tend to try to confuse students into learning a dozen ways to solve this same problem then force the students to use all the methods. Guess what? It just confuses the fuc& out of the students. I will show you one way, and it will always work (even if the dozen other methods may be easier in some cases).
[*]Choose any equation out of the three above. I will choose the first one: 0 = c + d*2.5+ e*6.25. You could choose any of the other equations, it will always work.
[*]I simplified it a bit by multiplying 2.5*2.5 = 6.25. Always simplify if possible. It'll keep your problem clean, managable, and less confusing.
[*]Rearrange to get ANY of the variables alone on one side. I will choose "c". A little subtraction gives you this:

• c = - d*2.5 - e*6.25.

[*]Do you understand how I got this?
[*]Bingo! I already know what "c" is equal to. All I have to do is find "d" and "e" and I'll know "c".

**Plug "c" into either of the other equations.
[*]Pick either equation #2 or #3. Remember we already used #1, so forget it ever existed. I'll pick equation #2: 1 = c + d*4+ e*4*4.
[*]We know what c is equal to. I bolded it above. Thus plug it into equation #2.

• 1 = (- d*2.5 - e*6.25) + d*4+ e*4*4.

[*]Do you see where I put what "c" was equal to into that equation?
[*]Simplify to make life easier. I will combine "d*4" and "-d*2.5" to be "d*1.5". Also I will combine the "e" terms. The final result is:

• 1 = d*1.5 + e*9.75.

Repeat the steps marked with ** above.
[*]Solve for "d" or "e". I'll choose "d"

• d*1.5 = 1 - e*9.75.
  d = (1 - e*9.75)/1.5.
  d = 2/3 - e*6.5.

[*]Bingo! Now we know what "d" is! Of course, unfortunately we just have to first find "e".
[*]Do you remember where above I said we can solve for "c" if we only knew "d" and "e"? Well, now we know "d". Lets simplify "c":

• c = - d*2.5 - e*6.25 = -(2/3 - e*6.5)*2.5 - e*6.25.

[*]Do you see where I plugged in the formula for "d"?
[*]That is a nasty formula, I will simplify it (I'll let you try this part on your own). The simplified version is:

• c = - 5/3 + e*10.

[*]Look! Now all we need to do is to find "e" and we'd know both "c" and "d".

[*]Plug "c" and "d" into the third and final equation: 2 = c + d*5+ e*5*5.

• 2 = (- 5/3 + e*10) + (2/3 - e*6.5)*5 + e*25.

[*]Do you see how I plugged those in?
[*]Oh crap! This is complicated again. I always say, simplify to make life easier. Do a little algebra (I'll let you do this on your own) and you get:

• 2 = 5/3 + e*2.5.

[*]Solve for "e":

• e*2.5 = 2 - 5/3 = 1/3.
  e = (1/3)/2.5 = 1 / 7.5 = 0.133333333.

Find "c" and "d":
[*]We needed to know "e" to find both "c" and "d". Since we now know "e", we can do it all.

• c= - 5/3 + e*10 = -5/3 + 10/7.5 = -1/3 = -.333333333333.
  d = 2/3 - e*6.5 = 2/3 - 6.5/7.5 = -1/5 = -0.2.

We know it all. Plug, "c", "d", and "e" into the original equation:

• B = -1/3 + -0.2*A + 1/7.5

Problem solved. This looks long and complicated, but it can be done in under 5 minutes with some practice. With Excel, it can be done in under 30 seconds. Jedec started it, I'll show you the Excel result. Click me!

 

[dopcombo]

perhaps if its homework, he should just type out exactly the question as it has been phrased by the teacher.

because the way he explains it, it could be anything.

edit: oh wait i just saw you say you're 25 in another thread. then i have no clue how to help you. Could you maybe give more context? such as why or how you got those 3 data points to begin with?

 

[Accipiter22]

thanks guys! That'll help on the project I'm working on a ton....I can actually use what you guys gave me and apply it to several other issues I'm having, thanks for showing me how to do substitution...as a reward I'll no longer bog the servers down with YAGT's