Polynomial function graphs have similarities depending on their degree. Explain how you can determine the best regression model by using finite differences. Then determine the end behavior of the graph based on the degree of a function and information gathered from a data table.

Question
Exponential growth and decay
asked 2020-11-08
Polynomial function graphs have similarities depending on their degree. Explain how you can determine the best regression model by using finite differences. Then determine the end behavior of the graph based on the degree of a function and information gathered from a data table.

Answers (1)

2020-11-09
Step 1
Determining the best regression model by using finite differences. The method of finite differences is used for curve fitting with polynomial models. It is best introduced by an example as shown in table given below. For the sequence of x value inputs at the regularly spaced points \(\displaystyle-{3},-{2},-{1},\cdots\ {3},\) the y value outputs are shown in the table.
Table 1 is given below.
From the table, we can understand that this is third difference finite function of third degree.
This explains
\(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}.\)
Step 2
Here we need to find the value of a, b, c and d
\(\displaystyle{a}={\frac{{-{6}}}{{{3}!}}}\)
\(\displaystyle{a}={\frac{{-{6}}}{{{6}}}}\)
\(\displaystyle{a}=-{1}\)
Step 3
Taking \(\displaystyle{x}={0},-{1},{1}\) for calculating the value of rest of the co-efficients.
\(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}\)
\(\displaystyle{f{{\left(-{1}\right)}}}={1}+{b}-{c}+{6}\)
\(\displaystyle{x}=-{1}\)
\(\displaystyle{y}={0}\)
\(\displaystyle{0}={b}-{c}+{7}\)
\(\displaystyle-{7}={\left({b}-{c}\right)}\)
\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{y}&\text{First Difference}&\text{Second Difference}&\text{Third Difference}\backslash{h}{l}\in{e}-{3}&{0}&\text{EMPTY VALUE}&\text{EMPTY VALUE}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}-{2}&-{4}&-{4}-{0}=-{4}&\text{EMPTY VALUE}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}-{1}&{0}&{0}-{\left(-{4}\right)}={4}&{4}-{\left(-{4}\right)}={8}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}{0}&{6}&{6}-{0}={6}&{2}&{2}-{8}=-{6}\backslash{h}{l}\in{e}{1}&{8}&{8}-{6}={2}&-{4}&-{6}\backslash{h}{l}\in{e}{2}&{0}&{0}-{8}=-{8}&-{10}&-{6}\backslash{h}{l}\in{e}{3}&-{24}&-{24}-{0}={24}&-{16}&-{6}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)
Also
\(\displaystyle{f{{\left({1}\right)}}}=\ -{1}+{b}+{c}+{6}\)
\(\displaystyle{8}={b}+{c}+{5}\)
\(\displaystyle{b}+{c}={3}\)
Moreover,
\(\displaystyle{f{{\left({0}\right)}}}=-{0}+{0}+{0}+{d}\)
\(\displaystyle{x}={0}\)
\(\displaystyle{y}={6}\)
\(\displaystyle{6}={d}\)
Thus, finding b and c by subtracting and adding equation (1) and (2)
\(\displaystyle{2}{b}=-{4}\)
\(\displaystyle{b}=-{2}\)
\(\displaystyle{2}{c}={10}\)
\(\displaystyle{c}={5}\)
At last the finite function becomes:
\(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}-{2}{x}^{{{2}}}+{5}{x}+{6}\)
Step 4
Sketching a grph from the function found above
Thus, end behaviour of \(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}-{2}{x}^{{{2}}}+{5}{x}+{6}\)
is described from the table as below
As \(\displaystyle\rightarrow\infty,\ {f{{\left({x}\right)}}}\rightarrow\ -\infty\) and As \(\displaystyle{x}\rightarrow\ -\infty,{f{{\left({x}\right)}}}\rightarrow\infty\)
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Source:
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1
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