Oh, use a graphing calculator program! I think you can download a trial version free. And you'll never look back!

here are your three equations:

6=a+b+c
21=4a-2b+c
26=9a+3b+c

subtract 2 from 3 to get:

5=5a+5b

divide by 5 on both sides:

1=a+b => b=1-a

sub in 1-a for b in equation 2:

21=4a-2+2a+5 => 18=6a => a=3 (per the book)

to get c:

6=1+c => c =5

to get b:

6=3+b+5 => b=-2

y=3x^2-2x+5

QED

Oh, wow, just realized what I did wrong. It was a dumbass mistake. Sorry for bothering everyone.

HAHA Fuck you, CM

Sorry. But hey, I typed in almost all of my work before I realized my mistake (For someone reason I only multiplied one side of the = and not the other).

Hey FU abgemacht, you beat me to a math problem on this forum.

haha

I'd say we're even, but the last one was much harder, so you're still beating me.

Meh...single var calc is peanuts, I'd say we're pretty even, but I'm ∂ points ahead because I had to use limits.

lawl fair enough

using the forum for hw help. . . hmmmm so i need help on an Econ paper i'm writing for my Econ 321 class. So if someone could whip up a 15 to 20 pager on whether or not Indonesia will help or hurt the chances of Pacific Asia becoming the next econmic super region. also i have my name written down, so good luck ;)

FU both, I was sleeping.

Heh, I forget just how many mathematicians there are on here. Not a very interesting problem, but at least solvable with our knowledge.
A few of us just spent about an hour trying to justify something only to find out the man who proved it became very famous for it and thus it was well beyond us!

FUCKING EASY CHUMPS.

My first homework in university had us try to prove Fermat's little theorem, without telling us what it is. Guess who spent 10 hours trying to prove it, only to find the proof had a wikipedia article...

Quit using dip forums to do your homework for you. ;)

I disagree, it's certainly more productive than discussing religion or politics.

@ava

agreed

Plus, people here at least have some minimum intelligence, unlike Yahoo! Answers, where I also spend some time.

Chill out Furball!
@ava - Nice. Whenever I see homework like that I always think they should add "You may not ask for help or consult references - this must be your own work" then fail people who actually solve it, because they're either geniuses or cheated.

HAHAH

"I'm sorry, you're too smart, so you fail."

Mathcad 15 will solve all your problems :)

touche, ava... touche.

Mathcad won't solve my problems...

How do you come to those first three formulas? Can you explain that problem again, more slowly? I had the same sorts of problems and I don't understand them at all.

Are you being serious? If so, I'd be happy to help. But, I don't want to waste time if you're joking.

You just plug the x and y coordinates into the quadratic expression. Three points, so three equations.

I am being perfectly serious. My skills don't lie in the realm of mathematics, as fascinating as it can be at the higher levels.

How do you know which coordinate goes where, then? And how?

In general, a point is represented by (x,y). So, the first point (1,6) means that x=1 and y=6.

we have the equation y=ax^2+bx+c. So, we plug in 1 for x and 6 for y to get:

6=a(1)^2+b(1)+c

Finally, this gives us:
6=a+b+c

The same can be done for the other points.

To be more correct, I should say that the point in a 2 rectangular coordinate system (which you are most used to) is represented (x,y).

If it was 3 , you would have points like (2,10,7), which would correspond to (x,y,z).

Alright, so then you would plug each set of coordinates, and then how does one come to the formula of the parabole? Or is that already given?

That is usually given. Equations for parabolas, hyperbolas, ellipses, are well known are are rather tricky to determine by yourself.

In the real world, of course, you often don't have a formula. Instead, you have tons of points and you're trying to get an approx. of the formula.

Or, you have a formula, but it can't be solved analytically (the one in this example can be), so you need to "massage" the formula or solve numerically.

If I may ask, why would you assume that I wasn't being serious? Are you so cynical? ;-)

And thanks for taking the time to explain it a bit, by the way.

I didn't assume you weren't being serious. If I had, I wouldn't have responded. But, it was hard to tell, especially with other people in this thread already complaining about how easy the problem was.

Easy is a relative term. I don't have a great interest in math, so that makes it more difficult to do well in class. Thank you for your help!

Happy to help.

And, you should try to pay attention. Math is pretty awesome, but you won't be able to do the cool stuff if you can't do the basics.

I do pay attention, usually, but I doubt I'd ever study mathematics. While I'm sure it's fascinating, it takes far too long for me to understand the basics, and by then the rest of the class understands it and has moved on. I much prefer playing with words than with numbers anyway.

STUDY MATHS!!
I'd be interested to know how many people on here actually study mathematics at college/uni/are maths alumni - maybe I should start a thread for it somewhere.

re "How do you know the formula":
There are generic formulae for a number of common graphs, circle, ellipse, parabola, hyperbola.

it's odd how 'how do you know the equation of a parabola' isnt something i thought of as a possible question...

I rate this problem a 14, as in that's when i learned to solve it... of course i could also have rated it at [Insert current age here] as i forgot how to solves it at about the age of 17...

I've been doing those steps automatically using a computer to fit the line to the points. (exactly, or a splice to get a nice line which fits together...)

But i haven't done it by hand in soo long, i didn't have a clue where the three simultanoues equations came from.

A splice by the way is very cool (though i don't know how useful) it is where you have more than three points, and you fit a parabola through the first two and the second and third, but you need a third equation for your three variables... so you use the slope of the two parabolas (you require the first derivative of the first parabola to equal that of the second parabola)

That way your two parabolas will have the same slope at the point where they join...

you also want the second derivative to match to make it look like a nice join... hence a splice... I think i wrote some code to fit 8 of data with any order line, and any number of splices... lots of fun stuff...

@Figle: When I go to college (in 4 years) I'm probably going to major in mathematics.

@Figle

I have a minor in math (3 or 4 courses short of a major)

@Ora - I assume that would be less efficient than just substituting in as we did above and then solving the linear equations by reduction?

@figles, huh?

i'm thinking of getting a nice curve going through several points.

two points defines a straight line - or many parabola's which could pass through them (as you have one free variable)
three points defines one parabola (quadratic line)

So 8 defines a x^7 power equation which is a bit crazy... but is done the same way - by subbing into the equations...

Instead you can fit several parabolas together, (onto two points each, or maybe three poins for the first one... i'm not sure) and match their slope and chang in slope... nothing to do with being efficient, it just looks like you have a smooth curve (instead of a high order equation which describes the whole thing... but looks nothing like what you'd expect to see in real life...)

Not that a splice is particularily correct or useful, but it's a cool way of drawing a line through points...