For each of the following tables of data, determine whether the data is linear, quadratic or neither. Please explain. Thank you!

- KJ4UTS

- jamiebookeater

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- KJ4UTS

##### 1 Attachment

- KJ4UTS

2.1204-1.5203 = 0.6001
2.9574-2.1204 = 0.837
4.1248-2.9574 = 1.1674
5.7531-4.1248 = 1.6283

- KJ4UTS

##### 1 Attachment

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- KJ4UTS

4.454-2.342 = 2.112
6.566-4.454 = 2.112
8.678-6.566 = 2.112
10.790-8.678 = 2.112

- KJ4UTS

##### 1 Attachment

- KJ4UTS

3.128-1.232 = 1.896
5.888-3.128 = 2.76
9.512-5.888 = 3.624
14.0-9.512 = 4.488

- KJ4UTS

How do I know if its linear, quadratic or neither, I think I did the work right?

- KJ4UTS

I think the second one is linear?

- FireKat97

I was thinking along the same lines as you... tbh Im not too sure about this one either :/

- KJ4UTS

I think I got it the second one is linear because the y differences equal and the third one is quadratic because the 2nd differences equal. I have not checked the first one yet.

- KJ4UTS

The first one looks like neither linear nor quadratic.

- anonymous

Linear is when the difference between the points are all the same.
Quadratic is when the difference between the points are increasing or deceasing each time.
Neither is when it increases slower or decreases slower each time.

- anonymous

For me number one is quadratic.

- anonymous

Quadratic equation consist of parabola
|dw:1442894201527:dw|
The slopes increase in a parabola.

- anonymous

Just like the slopes of number one.

- anonymous

Linear is y=mx+b
Let m be 5 and b be 2
(0,2) (1,7) (2,12)
The difference is 5 for each one.

- anonymous

Nice job attempting it though.

- KJ4UTS

Thank you!
I turned in the assignment and going from top to bottom:
1. Neither
2. Linear
3. Quadratic

- anonymous

Does it have to be one of each?

- KJ4UTS

For number 1 if you subtract the seconds you get different results:
2.1204-1.5203 = 0.6001
2.9574-2.1204 = 0.837
4.1248-2.9574 = 1.1674
5.7531-4.1248 = 1.6283
0.837-0.6001= 0.2369
1.1674-0.837=0.3304
1.16283-1.1674=0.4609
All different means its neither.
For number 2 all the y's are the same:
4.454-2.342 = 2.112
6.566-4.454 = 2.112
8.678-6.566 = 2.112
10.790-8.678 = 2.112
Making it linear.
For number 3 if you subtract the 2nds you get 18.4 for all of them:
3.128-1.232 = 1.896
5.888-3.128 = 2.76
9.512-5.888 = 3.624
14.0-9.512 = 4.488
4.488-3.624 = 18.4
3.624-2.76 = 18.4
2.76-1.896 = 18.4
2nds all the same mean its quadratic.

- anonymous

4.488-3.624 = 18.4
3.624-2.76 = 18.4
2.76-1.896 = 18.4
Its 0.864

- anonymous

Yes you are correct.
Sorry for mentioning the wrong one on number one.

- anonymous

I didnt calculate the change of slope because I didnt have a calculator with me.
1. somewhat looks quadratic without a calculator lol.

- KJ4UTS

No problem thanks for your time and help.
Thank you for catching me on that mistake, the answer was 18.4 was for a question and I got mixed up on all the work I did on paper:
Consider the quadratic function f(x)=2.3x^2+5.2x-6.7 with the input values x = 2, 4, 6, 8 and 10 and I had to determine the constant value of the second differences and it was 18.4

- KJ4UTS

I am just a little tired I guess lol, have a great night :)

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