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#1 2009-08-16 18:27:16

Greaterpathmagician
Member
Registered: 2009-06-16
Posts: 32

Finding the quadratic equation

Hello

I don't know how to find the quadratic equation from a table of values.

Here is the question and some of the tables of values.

Question: For those that are one of the above five types determine the algebraic rule for the reationship in the form of "y=???'.
("udef" indicates that the function is undefined for that value of x)

a) x: -4, -3, -2, -1, 0, 1, 2, 3, 4
    y:  12, 6,  2, 0,  0, 2, 6, 12, 20

b) x: -4, -3, -2, -1, 0, 1, 2, 3, 4
    y:  6, 2, 0, 0, 2, 6, 12, 20, 30

I know how to identify if it is quadratic (i.e the second difference) but not how to find and write the equation.

Thank you

Last edited by Greaterpathmagician (2009-08-16 18:29:30)

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#2 2009-08-16 18:39:42

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 89,150

Re: Finding the quadratic equation

Hi Greaterpathmagician;

a) y=x^2+x

b) y= x^2+3x+2

These were found by interpolation and there are other ways.

Last edited by bobbym (2009-08-16 18:44:20)


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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#3 2009-08-16 23:27:48

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Finding the quadratic equation

Second difference is one way to do it.

6, 2, 0, 0, 2, 6, 12, 20, 30

-4, -2, 0, 2, 4, 6, 8, 10

2, 2, 2, 2, 2, 2, 2

The second difference is 2, and dividing that by 2 gets you the coefficient of the quadratic term.
So now you know that the equation will be x², plus a linear part.

One way of finding the linear part is to take x² away from the terms in the sequence and analyse what's left.

-10, -7, -4, -1, 2, 5, 8, 11, 14


Why did the vector cross the road?
It wanted to be normal.

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