Determine the algebraic modeling which of the following data sets are linear and which are exponential. For the linear sets, determine the slope.

Determine the algebraic modeling which of the following data sets are linear and which are exponential. For the linear sets, determine the slope.

Question
Modeling
asked 2021-02-11

Determine the algebraic modeling which of the following data sets are linear and which are exponential. For the linear sets, determine the slope. For the exponential sets, determine the growth factor or the decay factor
a) \(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & \frac{1}{9} & \frac{1}{3} & 1 & 3 & 9 & 27 & 81 \\ \hline \end{array}\)

b) \(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 2 & 2.6 & 3.2 & 3.8 & 4.4 & 5.0 & 5.6 \\ \hline \end{array}\)

c) \(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 3.00 & 5.0 & 7 & 9 & 11 & 13 & 15 \\ \hline \end{array}\)

d) \(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 5.25 & 2.1 & 0.84 & 0.336 & 0.1344 & 0.5376 & 0.021504 \\ \hline \end{array}\)

Answers (1)

2021-02-12
Part (a)
Given data set is
\(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & \frac{1}{9} & \frac{1}{3} & 1 & 3 & 9 & 27 & 81 \\ \hline \end{array}\)
\(\displaystyle\therefore\ \displaystyle\frac{{{y}{\left({1}\right)}}}{{{y}{\left({0}\right)}}}=\frac{3}{{1}}={3}\ \text{and}\ \displaystyle\frac{{{y}{\left({3}\right)}}}{{{y}{\left({2}\right)}}}=\frac{27}{{9}}={3}\)
Hence given data set is exponential.
Since y is increasing, so model is growth model.
Growth factor\(=3\)
Part b)
Given data set is
\(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 2 & 2.6 & 3.2 & 3.8 & 4.4 & 5.0 & 5.6 \\ \hline \end{array}\)
\(\displaystyle\therefore{y}{\left({1}\right)}-{y}{\left({0}\right)}={0.6}{\quad\text{and}\quad}{y}{\left({2}\right)}-{y}{\left({1}\right)}={0.6}\)
Hence data set is linear.
Slope \(\displaystyle=\frac{{{y}{\left({1}\right)}-{y}{\left({0}\right)}}}{{{1}-{0}}}=\frac{{{3.8}-{3.2}}}{{1}}={0.6}\)
Part c)
Given data set is
\(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 3.00 & 5.0 & 7 & 9 & 11 & 13 & 15 \\ \hline \end{array}\)
\(\displaystyle\therefore{y}{\left({1}\right)}-{y}{\left({0}\right)}={2}{\quad\text{and}\quad}{y}{\left({2}\right)}-{y}{\left({1}\right)}={2}\)
Hence data set is linear.
Slope \(\displaystyle=\frac{{{y}{\left({1}\right)}-{y}{\left({0}\right)}}}{{{1}-{0}}}=\frac{{{9}-{7}}}{{1}}={2}\)
Part d)
Given data set is
\(\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 5.25 & 2.1 & 0.84 & 0.336 & 0.1344 & 0.5376 & 0.021504 \\ \hline \end{array}\)
\(\displaystyle\therefore\frac{{{y}{\left({1}\right)}}}{{{y}{\left({0}\right)}}}=\frac{0.336}{{0.84}}={0.4}\)
Hence given data set is exponential.
Since y is decreasing, so model is decay model.
Decay factor \(= 0.4\)
0

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