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Mathematics 9 Online
OpenStudy (anonymous):

Use successive differences to classify the function represented in the table. Show all your work.

OpenStudy (anonymous):

OpenStudy (anonymous):

Here's what I know: The line is NOT linear since there is no pattern. It is quadratic though.

OpenStudy (anonymous):

same as computing a bunch of slopes only the denomintor is always so just the difference in the numerators −10+29=19 −3+10=7 −2+3=1 −1+2=1 6+1=7 25−6=19 the recessive differences are quadratic, to the original function is "other" namely cubic y=x^3−2 will work if you try the pairs

OpenStudy (anonymous):

I googled it already. It is not the same problem.

OpenStudy (anonymous):

@misty1212 @texaschic101 @amistre64

OpenStudy (anonymous):

@e.mccormick @iambatman

OpenStudy (e.mccormick):

@helper99 umm... the y axis values are all positive, so it can't be cubic. @TheUnbreakable have you tried fitting a parabola to it?

OpenStudy (anonymous):

I am not even sure what a parabola is..

OpenStudy (anonymous):

can you refresh my memory please? I want to learn this...

OpenStudy (e.mccormick):

Quadratics make parabolic curves: |dw:1429641913146:dw| and things similar to that.

OpenStudy (e.mccormick):

But smoother... hehe.

OpenStudy (anonymous):

:) I know what you are trying to show me. So how do I apply this "parabola" to my problem? It sounds like a distant cousin of the ebola virus....

OpenStudy (e.mccormick):

Well, quadratics graph as one. What have you done in graphing quadratics?

OpenStudy (anonymous):

I know the formula. That is about it... Sorry, I am really forgetful....

OpenStudy (e.mccormick):

OK. Here are two graphs just to show more clearly what I am talking about: https://www.desmos.com/calculator/uxoyz3bvs9 So the question becomes, how do you find a quadratic from points.

OpenStudy (anonymous):

Would the slope help? All I know if these types of lines are infinite and have an infinite amount of points.

OpenStudy (e.mccormick):

Slope does not apply in the case of curves until you know calculus. Now, a quadratic can be witten more generally as: \(y=ax^2+bx+c\) So you need to find a, b, and c. Well, to find three unknowns you need 3 equations.

OpenStudy (anonymous):

Alright, so where do I look for these three unknowns?

OpenStudy (amistre64):

use the difference tiers ....

OpenStudy (anonymous):

@amistre64 I did try. The differences are not in a consistent pattern.

OpenStudy (amistre64):

14 5 2 5 14 whats our first set of differences? humor me

OpenStudy (e.mccormick):

Well, you are given the x and y values. the y is the h(x). So you know things like this \(y=ax^2+bx+c\) becomes: \(2=a0^2+b0+c\)

OpenStudy (anonymous):

@amistre64 -9, -3, 3, 9

OpenStudy (amistre64):

14 5 2 5 14 -9 -3 3 9 and whats the differences in the difference?

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