Deriving the formula from a graph

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Discussion Overview

The discussion revolves around deriving a formula from a graph that represents a sequence of points, specifically examining the relationship between the x and y coordinates. The focus is on identifying a general formula that fits the observed values, which appear to follow a geometric progression.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant presents a set of points and suggests that the y-values follow a geometric sequence, proposing a formula based on this observation.
  • Another participant identifies the pairs as ##(n, 2^{n+1})## for ##n=0,1,2,...##, implying a specific relationship between x and y.
  • A question is raised about how the second participant derived their formula and whether the general formula could be ##y=2^{x+1}##.
  • There is a reiteration of the observation that the y-values are powers of 2, suggesting that adjusting the exponent aligns the y-values with the x-values.

Areas of Agreement / Disagreement

Participants express differing views on the correct general formula, with some proposing ##y=2^{x+1}## while others suggest alternative formulations. The discussion remains unresolved regarding the definitive general formula.

Contextual Notes

There are limitations in the assumptions made about the starting point of x-values and the interpretation of the graph, which may affect the proposed formulas.

adjacent
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So,here we have a graph.For every unit of 1(x axis),the unit of the previous y-axis is multiplied.
Like this.
(0,2),(1,4),(2,8),(3,16)...
I don't know how the formula will be,and had never seen this kind of graph.
The y seems to be a geometric sequence,on which we can use, ## ar(n-1) ## .r is 2.a is 2
So 2 x 2(n-1)
n is the value of x.For every value of x,we can now find the value of y.
So how can we find the general formula of the graph?Is it ## y=2 x 2(x-1) ## ? I don't think so,If the x values started with 1,it's possible.

I don't know why LaTeX is not working,Help please.
 
Last edited:
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Your pairs are ##(n, 2^{n+1})## for ##n=0,1,2,...##.
 
How did you derive that?What could be the general formula? ##y=2^{x+1}## ?
 
LCKurtz said:
Your pairs are ##(n, 2^{n+1})## for ##n=0,1,2,...##.

adjacent said:
How did you derive that?What could be the general formula? ##y=2^{x+1}## ?

Yes. You already noticed the ##y## values were powers of ##2##. So it was just a matter of adjusting the exponent to make the ##y## value agree with the ##x## value.
 
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Thanks, (lol I take reliable and cheap)
 

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