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2.1204-1.5203 = 0.6001 2.9574-2.1204 = 0.837 4.1248-2.9574 = 1.1674 5.7531-4.1248 = 1.6283
4.454-2.342 = 2.112 6.566-4.454 = 2.112 8.678-6.566 = 2.112 10.790-8.678 = 2.112
3.128-1.232 = 1.896 5.888-3.128 = 2.76 9.512-5.888 = 3.624 14.0-9.512 = 4.488
How do I know if its linear, quadratic or neither, I think I did the work right?
I think the second one is linear?
I was thinking along the same lines as you... tbh Im not too sure about this one either :/
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.
The first one looks like neither linear nor quadratic.
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.
For me number one is quadratic.
Quadratic equation consist of parabola |dw:1442894201527:dw| The slopes increase in a parabola.
Just like the slopes of number one.
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.
Nice job attempting it though.
Thank you! I turned in the assignment and going from top to bottom: 1. Neither 2. Linear 3. Quadratic
Does it have to be one of each?
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.
4.488-3.624 = 18.4 3.624-2.76 = 18.4 2.76-1.896 = 18.4 Its 0.864
Yes you are correct. Sorry for mentioning the wrong one on number one.
I didnt calculate the change of slope because I didnt have a calculator with me. 1. somewhat looks quadratic without a calculator lol.
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
I am just a little tired I guess lol, have a great night :)