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 \(= 0.4\)

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 \(= 0.4\)