How to find function of graph based on 150 points, mins, maxs?

A

anonymouscoward

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I've been trying to figure out how to find the function of a particular graph based on a set of points. This would be easy regularly, but the equation is kinda curved. I should've payed more attention in math class, and that's probably my fault, but I've been searching for how to find the function of this graph forever and I really need help... I've already figured out that each point on the graph seems to be the point before it times .95 or something around that. The graph shows a curve that is gradually decreasing. There are a total of 150 points on the graph. Here is some information I know is correct:
* Lowest X Value: 0
* Highest X Value: 150
* Lowest Y Value: 0.0455554974483661
* Highest Y Value: 100
Also, here is copy of the graph (Click for Full Size):

Can anybody help me find the function of this? I've been trying to figure this out for a long time... also, I'm not sure this is the right forum, so if not, please move it and inform me. Thanks for all your help!
 
Re: How do I find the function of a graph based on points?

anonymouscoward said:
I've been trying to figure out how to find the function of a particular graph based on a set of points. This would be easy regularly, but the equation is kinda curved. I should've payed more attention in math class, and that's probably my fault, but I've been searching for how to find the function of this graph forever and I really need help... I've already figured out that each point on the graph seems to be the point before it times .95 or something around that. The graph shows a curve that is gradually decreasing. There are a total of 150 points on the graph. Here is some information I know is correct:
* Lowest X Value: 0
* Highest X Value: 150
* Lowest Y Value: 0.0455554974483661
* Highest Y Value: 100
Also, here is copy of the graph (Click for Full Size):

Can anybody help me find the function of this? I've been trying to figure this out for a long time... also, I'm not sure this is the right forum, so if not, please move it and inform me. Thanks for all your help!
Click to expand...

This is an "exponential" curve fitted into

\(\displaystyle y \, = \, A\cdot B^{- x}\)

Take log of both side and the curve will become linear - and you can find the constants.

To brush up methods to find best-fit-straight line do a google search.
 
Exponential decay ...

Hi AC:

The graph is an example of exponential decay.

\(\displaystyle f(x) \;=\; a \cdot e^{k \cdot x}\)

a is a constant

e is the base of the natural logarithm

k is a constant

x is the variable

When k is a positive number, then this function models exponential growth (like interest growth on an investment). When k is a negative number, then this function models exponential decline (like the decay of a chunk of radioactive stuff).

Your reasoning that the slope between a pair of points is -0.95 won't hold throughout the domain of the graph. This is because the slope in exponential functions is determined by adjacent values. In other words, when y is big, the change in slope is big going from point to point. When the y values get small, then the change in slope is small.

This is the hallmark of exponential behavior. Your graph shows it. The decline is fast and furious at first (x = 0 .. 20), but the decline eventually slows to where the change is hard to see (x = 120 .. 145).

There are two methods to find a function to go with this graph. One method is straightforward; the other is not.

The easier method produces a graph that is similar to your data.

The more involved method produces a graph that is more precise.

Why are you doing this? Is this a homework assignment? If so, then what class is it from? Do you need a high level of precision?

For example, I can use the points (0, 100) and (10, 60) to get the following function.

\(\displaystyle f(x) \;=\; 100 \cdot e^{-0.05108 \cdot x}\)

This function models your data to some extent, but it may not be precise enough for the larger values of x.

\(\displaystyle f(75) \;=\; 2.2\)

From your graph, it appears to me that when x is 75, y should be about 0.5; however, as you can see above, the function outputs 2.2, instead.

To get a more precise function, we would need to use several data points and go through an exponential regression progress (more involved).

If you can give us an idea of why you're seeking a function, then we will have a better idea on how to help you.

Cheers,

~ Mark :)


 
Re: Exponential decay ...

mmm4444bot said:
Why are you doing this? Is this a homework assignment? If so, then what class is it from? Do you need a high level of precision?
Click to expand...
Weekend homework I had since wednesday to do it, and it's worth like 60 points... The function is supposed to be precise, probably around having 3 decimal places accurate when it is graphed...
 
Re: Exponential decay ...

anonymouscoward said:
The function is supposed to be precise, probably around having 3 decimal places accurate when it is graphed
Click to expand...

This level of precision concerns me. Did you come up with the value of "3 decimal places" yourself? I'm asking because you wrote, "probably".

(Graphing to within 1/1000ths of a unit is something only a machine can do!)

Anyway, to get that level of precision, you would need to have data to match.

Do you have a table of data? Did you get anything other than the graph?

For example, from the graph I see a point (10, 60).

But perhaps you have the data to show that the actual values are something like (9.996, 60.012).

What techniques is your class teaching? Are you studying regression methods?

Is your class using any computer software or graphing calculator?

Please respond to the questions in blue.

~ Mark :)



 
Re: Exponential decay ...

Did you come up with the value of "3 decimal places" yourself?
Click to expand...
The data given to me was rounded to 3 decimal places. Anyways, here are the points on the graph, separated by spaces, as I can't code tables:
X Values Y Values
0 100.000
1 95.000
2 90.250
3 85.738
4 81.451
5 77.378
6 73.509
7 69.834
8 66.342
9 63.025
10 59.874
11 56.880
12 54.036
13 51.334
14 48.767
15 46.329
16 44.013
17 41.812
18 39.721
19 37.735
20 35.849
21 34.056
22 32.353
23 30.736
24 29.199
25 27.739
26 26.352
27 25.034
28 23.783
29 22.594
30 21.464
31 20.391
32 19.371
33 18.403
34 17.482
35 16.608
36 15.778
37 14.989
38 14.240
39 13.528
40 12.851
41 12.209
42 11.598
43 11.018
44 10.467
45 9.944
46 9.447
47 8.974
48 8.526
49 8.099
50 7.694
51 7.310
52 6.944
53 6.597
54 6.267
55 5.954
56 5.656
57 5.373
58 5.105
59 4.849
60 4.607
61 4.377
62 4.158
63 3.950
64 3.752
65 3.565
66 3.387
67 3.217
68 3.056
69 2.904
70 2.758
71 2.620
72 2.489
73 2.365
74 2.247
75 2.134
76 2.028
77 1.926
78 1.830
79 1.738
80 1.652
81 1.569
82 1.491
83 1.416
84 1.345
85 1.278
86 1.214
87 1.153
88 1.096
89 1.041
90 0.989
91 0.939
92 0.892
93 0.848
94 0.805
95 0.765
96 0.727
97 0.691
98 0.656
99 0.623
100 0.592
101 0.562
102 0.534
103 0.508
104 0.482
105 0.458
106 0.435
107 0.413
108 0.393
109 0.373
110 0.354
111 0.337
112 0.320
113 0.304
114 0.289
115 0.274
116 0.261
117 0.248
118 0.235
119 0.223
120 0.212
121 0.202
122 0.192
123 0.182
124 0.173
125 0.164
126 0.156
127 0.148
128 0.141
129 0.134
130 0.127
131 0.121
132 0.115
133 0.109
134 0.104
135 0.098
136 0.093
137 0.089
138 0.084
139 0.080
140 0.076
141 0.072
142 0.069
143 0.065
144 0.062
145 0.059
146 0.056
147 0.053
148 0.050
149 0.048
150 0.046

That took a while to type, as I had it on a sheet that was printed out, split into 3 columns of 50... anyways... I think it starts at 100 and each one is multiplied by .95, but I didn't actually check for each one. The first like 20 seem to be so however. I don't exactly know what you mean by regression methods, but we have been working on graphing curves. I don't pay enough attention in class. :oops:
 
Re: How do I find the function of a graph based on points?

I finished checking all the points, and each one seems to be the previous one times .95, rounded to 3 decimal places. Now that I have that figured out, how would I make that into a function?
 
Re: Exponential decay ...

anonymouscoward said:
That took a while to type, as I had it on a sheet that was printed out ...

I regret that you felt a need to do that; a sample would have been fine.

... I don't exactly know what you mean by regression methods ...

I was not clear. Computers have built-in programs that fit data to a model. For you, the TI-83 will do the regression and provide numerical values for an equation that provides a "best fit" to the given data.

... we have been working on graphing curves ...
Click to expand...

I'm guessing that you are supposed to enter this data as a list into the graphing calculator, and use the calculator to come up with the parameters for a model (function).

Does that sound right?

Other than the data and the graph, did you get any written instructions?

Do you have the user manual for the TI-83? (I will look on-line at the user manual for the TI-83 to see what types of built-in regression programs it has.)

~ Mark :)
 
Re: How do I find the function of a graph based on points?

The TI-83 Plus has a built-in regression program (ExpReg). It will calculate from your data the values of A and B for the exponential function posted by Subhotosh.

This program is accessed from the STAT CALC menu. (See page 356 in the user guide.)

Do you have instructions for this exercise?

I don't know why you have a table of data. Sometimes, instructors use tables and graphs as teaching aids, since functions can be described in words, by data, by graphs, and by algebraic expressions.

There is no need to use the calculator unless you are sure that you need to get a best-fit function from that specific data.

If the instructions call for you to do something by hand, then that will be easier.

~ Mark :)


 
Re: How do I find the function of a graph based on points?

I entered the X Values as the List X1, and the Y Values as the list L2, then typed ExpReg L1,L2 and the calculator said:
Code: 
ExpReg
Y=a*b^x
a=100
b=.95
So that means the function for this graph is f(x)=100*.95^X?
 
Re: How do I find the function of a graph based on points?

Yep. You can also write it as \(\displaystyle 100\cdot (\frac{19}{20})^{x}\)

Looks nicer without the decimal.

You wouldn't have has to use all those points. I just used the first few and got the same result.

I used the EXPREG in my TI-92 as well.
 
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