is there another way of solving this kind of question

betahaus · 2010-01-25 10:21:13

right... i did a question earlier that said the equation of a curve is in the form y = ax^2 + bx + c. if it passes through the points (0,2), (2,3), and (5,12). find the values of a, b, and c, and thus find the equation of the curve.

i know that you can solve this using simultaneous equations, which i can do easily. my question is, is there another way to solve this problem (that is possibly easier?) in addition, how do i solve this problem with matrices? thx

John E. Franklin · 2010-01-25 10:47:38

@betahaus, can you post the min coordinate if you have it?
I have been doing trial and error calculations but didn't figure it out.
Is the min to the right or left of (0,2) ?

The squaring effect takes place based on the x-distances from the min coordinate, so I was guessing at the minimum point coordinate and then doing x-distance squaring and comparing that column of numbers to the y-distance observed, which would give you the "a" factor in ax^2, providing the min point was actually correct.
In my guesses, the min point was off, as the ratios were not consistent between each other.

Last edited by John E. Franklin (2010-01-25 11:31:00)

John E. Franklin · 2010-01-25 12:02:11

I relearned the 3 simultaneous equations way and solved for a, b, and c.
a = -b and c = 2 and a = 1/2.
I found the minimun point as (0.5, 1.875) with calculus.
Then I made this list:

0.5  .25   .125       This list is the absolute x-distance from minimum to (0,2), then the square of it (.25), and the y-distance, notice as half of .25 in middle column, due to a=1/2.
1.5  2.25   1.125  This is min point to (2,3)
4.5  20.25  10.125  This is min point to (5,12)

But working backwards may be hard to do. Not sure how to narrow in on it...

John E. Franklin · 2010-01-25 14:09:03

How about if you do another type of simultaneous equations, that solve for the minimum point instead of a, b, and c.
How about define minimum point as (m, n).
Then remember we have these 3 points:
(0,2), (2,3), and (5,12).
So now, the following maybe, just may be correct if I am making any sense:
(m-0)^2/(n-2) = (m-2)^2/(n-3) = (m-5)^2/(n-12)
And those ratios above are also "a" in the y = a x^2 + b x + c.

John E. Franklin · 2010-01-25 14:21:07

trying to solve... (n-3)m^2=(n-2)(m-2)^2 and (n-12)m^2 = (n-2)(m-5)^2
Two equations, two unknowns, maybe can do it...
n m^2 - 3 m^2 = (n-2)(m^2 - 4m + 4) and n m^2 - 12 m^2 = (n-2)(m^2 - 10m + 25)
ditto=nm^2-4mn+4n-2m^2+8m-8) and ditto2 = nm^2-10mn+25n - 2m^2 +20m - 50
n m^2 cancel on left and right sides...
0 = -4mn + 4n + m^2 +8m - 8 and 0 = -10mn + 25n + 10m^2 + 20m - 50
Hmm... Gettin' better...

John E. Franklin · 2010-01-25 14:28:36

try to get rid of the "mn" factor, that could pose a serious hardship I think.
Multiply the -4mn equation by -2.5 on both sides and add to the other equation to cancel the -10mn...
0 = 10mn - 10n -2.5m^2 - 20m + 20 and other equation is still the same (ends in - 50).
Now add them together.
0 = 15n + 7.5m^2 - 30
Woah, let me stop now and say I don't know how to solve this.
I'm going to cheat and see if the equation checks out though.
Let's plug in (0.5, 1.875) coordinate for minimum point just to see if we didn't make an error somewhere.
m = 0.5, n = 1.875
0 = 15(1.875) + 7.5(0.5)^2 - 30
0 = 28.125 + 1.875 - 30
Wow, magic! The equation is correct, I just can't solve it though!! Rats.

bobbym · 2010-01-25 16:13:58

Hi betahaus;

I took the liberty of renaming a,b,c to a1,b1,c1 so as not to interfere with the column vector b.

Now you solve Ax = b for x. If Ax = b were an ordinary equation you would divide by A and be done.
But Ax = b is a matrix equation and division is not defined for matrices.
But multiplication by the inverse of A is.

You get:

x = A^(-1) b

So a= 1/2 , b = -1/2 and c = 2.

Also I do not think there is any easier way to answer your top question
than by simultaneous equations. Especially since c is easy. What is simpler than a 2x2 set of linear simultaneous equations?

John E. Franklin · 2010-01-26 03:27:15

Tres bon! Bobby!
Does it matter which order you set up the A matrix?
Here are 2 choices?
Yours:

Or another one:

Can you point me to something to read to learn this better?

Last edited by John E. Franklin (2010-01-26 04:05:47)

bobbym · 2010-01-26 13:03:36

Hi John;

No, switching 2 columns has no effect on the answer: It would be as if you were solving this set:

c = 2
4a + 2b + c = 3
25a + 5b + c = 12

Or this set:

c = 2
2b + 4a + c = 3
5b + 25a + c = 12

You would get the same answer. This is all covered in Linear Algebra. Just pick up any text on it or any webpage
that is easy for you to understand. Post all your questions back here and I or someone
else will be glad to assist.

John E. Franklin · 2010-01-26 13:10:18

Thanks a lot bobby!
I took a course once, but
the teacher was really bad.

bobbym · 2010-01-26 13:45:22

Hi John;

Remember everything that was done in post #7 was just for show. It was just to demonstrate
the process. In reality you would never invert the A matrix, or do any of that.
You would use gaussian elimination or conjugate gradient, SOR or some other iterative
scheme to solve the set of n x n linear simultaneous equations.

John E. Franklin · 2010-01-26 14:56:10

Nice. Thanks so much for the information.

Math Is Fun Forum

#1 2010-01-25 10:21:13

is there another way of solving this kind of question

#2 2010-01-25 10:47:38

Re: is there another way of solving this kind of question

#3 2010-01-25 12:02:11

Re: is there another way of solving this kind of question

#4 2010-01-25 14:09:03

Re: is there another way of solving this kind of question

#5 2010-01-25 14:21:07

Re: is there another way of solving this kind of question

#6 2010-01-25 14:28:36

Re: is there another way of solving this kind of question

#7 2010-01-25 16:13:58

Re: is there another way of solving this kind of question

#8 2010-01-26 03:27:15

Re: is there another way of solving this kind of question

#9 2010-01-26 13:03:36

Re: is there another way of solving this kind of question

#10 2010-01-26 13:10:18

Re: is there another way of solving this kind of question

#11 2010-01-26 13:45:22

Re: is there another way of solving this kind of question

#12 2010-01-26 14:56:10

Re: is there another way of solving this kind of question

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