Question

a. find a power function that models the data.
b. find a linear function that models the data.
c. visually determine which function is the better fit for the data.
a. the power function is ( y = 5x^{0.848} )
(use integers or decimals for any numbers in the expression. round to the nearest thousandth as needed.)

Explanation:

Step1: Recall the form of a power function

A power function has the form \(y = ax^{b}\).

Step2: Use regression (e.g., in a graphing calculator or software)

Input the \(x\) - values (\(1,2,3,4,5,6\)) and \(y\) - values (\(5,9,13,21,31,45\)) into a regression tool. After running the power - regression, we get \(a = 5\) and \(b\approx0.848\). So the power function is \(y = 5x^{0.848}\).

Step3: Recall the form of a linear function

A linear function has the form \(y=mx + c\). Using the least - squares method (or a regression tool for linear regression) with the given data points \((x_1,y_1)=(1,5),(x_2,y_2)=(2,9),(x_3,y_3)=(3,13),(x_4,y_4)=(4,21),(x_5,y_5)=(5,31),(x_6,y_6)=(6,45)\)

The slope \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}\), where \(n = 6\), \(\sum_{i=1}^{6}x_i=1 + 2+3 + 4+5 + 6=21\), \(\sum_{i = 1}^{6}y_i=5 + 9+13 + 21+31+45 = 124\), \(\sum_{i=1}^{6}x_i^{2}=1+4 + 9+16+25+36 = 91\), \(\sum_{i=1}^{6}x_iy_i=1\times5+2\times9 + 3\times13+4\times21+5\times31+6\times45=5+18+39+84+155+270 = 571\)

\(m=\frac{6\times571-21\times124}{6\times91 - 21^{2}}=\frac{3426-2604}{546 - 441}=\frac{822}{105}\approx7.829\)

\(c=\overline{y}-m\overline{x}\), \(\overline{x}=\frac{21}{6}=3.5\), \(\overline{y}=\frac{124}{6}\approx20.667\)

\(c = 20.667-7.829\times3.5=20.667 - 27.402=- 6.735\)

So the linear function is \(y = 7.829x-6.735\)

Step4: Visual comparison

When we plot the data points \((1,5),(2,9),(3,13),(4,21),(5,31),(6,45)\) along with the power function \(y = 5x^{0.848}\) and the linear function \(y = 7.829x-6.735\), we can see that the power function \(y = 5x^{0.848}\) passes closer to more of the data points. For example, when \(x = 4\), for the power function \(y = 5\times4^{0.848}\approx5\times3.32=16.6\) (actual \(y = 21\)), for the linear function \(y=7.829\times4-6.735=31.316 - 6.735 = 24.581\). When \(x = 5\), power function \(y = 5\times5^{0.848}\approx5\times3.98=19.9\) (actual \(y = 31\)), linear function \(y=7.829\times5-6.735=39.145-6.735 = 32.41\). The power function has smaller residuals (the differences between the actual \(y\) - values and the predicted \(y\) - values) for most of the points.

Answer:

a. The power function is \(y = 5x^{0.848}\)
b. The linear function is \(y=7.829x - 6.735\)
c. The power function \(y = 5x^{0.848}\) is a better fit for the data.