Question

For the following exercises, enter the data from each table into MS-Excel and graph the resulting scatter plots (Use a smoothing function). Determine whether the data from the table could represent a function that is linear or not linear. n 1.25 2.25 3.56 4.2 5.65 6.75 7.25 8.6 9.25 10.5 f (n) 5.75 8.75 12.68 14.6 18.95 22.25 23.75 27.8 29.75 33.5  

I got the graph, but to how do I get to use the smoothing function to create another scatter plot graph?

The last question has two graphs in excel. The scatterplot and the exponential smoothing graph. I am currently still reading my math book but I want to know if how to get that exponential graph?

Answer 1

Given data points are -

n	1.25	2.25	3.56	4.2	5.65	6.75	7.25	8.6	9.25	10.5
f(n)	5.75	8.75	12.68	14.6	18.95	22.25	23.75	27.8	29.75	33.5

Let us check that these points can form a linear function or not .

We know that slope 'm' of a linear function given by -

f ni) – ያ(nj ከኪ ni ቦ

Here in the above table , we see that for any i and j the slope is always coming out as '3' ,i.e.,

$m = \frac{8.75-5.75}{2.25-1.25}= \frac{12.68-8.75}{3.56-2.25}= \cdots =\frac{33.5-29.75}{10.5-9.25}=3$

i.e., slope = m = 3 .

The equation of a linear graph is given by -

$f(x)=mx+c$

here , it will be -

$f(n)=mn+c$

where 'm' is the slope of the graph , and 'c' represents the y-intercept of the graph . W e already found out the value of 'm' , so to find 'c' we plug any of the given data set in the above equation .

$f(n)=mn+c$

Put (1.25 , 5.75 ) in above equation , we get -

$5.75=3(1.25)+c$

$\Rightarrow 5.75=3.75+c$

$\Rightarrow c=5.75-3.75=2$

Hence , we have the graph of above data set is given by -

$f(n)=3n+2$

SCATTERPLOT OF ABOVE GRAPH :

0 0 all u 0

NOTE:

Since the given data points represents a linear trend , therefore it would not include any exponential graph . An exponential graph would have been drawn if the given data sets showed us the linear trend .

Method to calculate function for exponential graph :

An exponential function is given by , f(x) = ab^x , where a represents the intercept and 'b' represents the common multiplier with which the function grows at equal interval .

To find 'b' : Choose any two values of 'x' from the given table of points , say x = x1 and x = x2 . Divide this interval from mid , so that we now have three points i.e., x1 , x0 , and x2 . Suppose 'b' is the common multiplier and we know the values of f(x1) and f(x2) , therefore we will find the value of 'b' in going from f(x1) to f(x2) , i.e.,

f(x1)*b*b = f(x2) , caculate 'b' from here and plug it in the exponential function f(x) .

To find 'a' = Choose any of the set ( x1, f(x1) ) or ( x2 , f(x2) ) in the function f(x) along with the value of 'b' and calculate the value of 'a' .

Hence , in this way , the required exponential function can be formulated by plugging the values of 'a' and 'b' in f(x) .