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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
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,430

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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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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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