tough equation

T

thecool

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Jul 14, 2009
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ok, if given that (x,y)and the coordinates are (1,2) (2,4) (3,6) i can figure out that the answer is y=x2 but my real dilemma is that my coordinates are (1,3) (2,1) and (3,2.5) i need a little help with the answer
 
The last set of data points, (1,3) (2,1) and (3,2.5), is another parabola with equation \(\displaystyle y=\frac{7}{4}x^{2}-\frac{29}{4}x+\frac{17}{2}\)

Do you have Excel or a nice calculator?. They will do these regressions and find the function which represents them.

I go the above equation with my calculator. That is the best way to do quadratic regressions.
 
thecool said:
ok, if given that (x,y)and the coordinates are (1,2) (2,4) (3,6) i can figure out that the answer is y=x2 but my real dilemma is that my coordinates are (1,3) (2,1) and (3,2.5) i need a little help with the answer
Click to expand...

You should state excatly what is NEEDED.

I GUESS, you need an equation of the curve running through the given points.

Try a quadratic equation of the form

y = Ax[sup:2zq36jhp]2[/sup:2zq36jhp] + Bx + C

You can find the values of A, B and C from the given points.

(1,3) ? 3 = A + B + C

(2,1) ? 1 = 4A + 2B + C

(3,2.5) ? 2.5 = 9A + 3B + C

Now you have 3 equations and 3 unknowns - solve those.
 
thanks for the help,but what i need is the one common equation that works for all three of the sets. like i said in my example the equation that worked for all three sets was y=x2. I need an equation that works for all three of the sets
 
You can find that by solving the system SK showed you. That is how you derive said equation.
 
Did you read our responses carefully???

You will get one equation (y = Ax[sup:39ipkrzp]2[/sup:39ipkrzp] + Bx + C).
 
ok jus a wee bit confused. not really working with any parabolas or Cartesian graphs, jus a regular set of numbers. but if thts the equation cool, thanks
 
thecool, are you understanding what is being said here?. To find the equation that describes the parabola when you are only given a set of data points one must derive it from the data points. That is why SK gave you that system of three equations with three unknowns. Solving that will give you the coefficients (A,B,C) you seek.

Solve the system of equation:

\(\displaystyle A+B+C=3\)

\(\displaystyle 4A+2B+C=1\)

\(\displaystyle 9A+3B+C=\frac{5}{2}\)

Solve this for A,B, and C and you have the equation you so desperately seek. It will be in the form \(\displaystyle Ax^{2}+Bx+C=0\)

I gave you the equation in the first post, so you know what you're shooting for. But I done it the easy way and used my calculator.
 
thecool said:
u lost me at quadratic
Click to expand...
C'mon, Cool; first, "u" is not a word; show you're serious by dropping chat-board lingo.

And are you too busy to look up "quadratic equation" using Google?
Not too interesting to help someone who shows no initiative...
 
Neither is C'mon. But anyhow, even tho i know the next numbers are (4,3.75) I still have no idea how to get to the equation
 
thecool said:
I still have no idea how to get to the equation
Click to expand...

thecool -- Be Cool ::

Dont you know how to solve the system of equation SK gave you , or you're not understanding the way he started up with ?

Be more specific .
 
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