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Find the power function that models the data in the table below.

x
1
2
3
4
5
6
7
8

y
3
4.6
5.7
7
8.1
9
10.2
10.3
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Part 1
The power function is yequalsnothing x Superscript nothing
  
  
.
​(Type integers or decimals rounded to three decimal places as​ needed.)

1 answer

To find a power function of the form \( y = ax^b \) that models the given data, we can use a method such as logarithmic transformation followed by linear regression.

Given the data:

\[ \begin{align*} x & : 1, 2, 3, 4, 5, 6, 7, 8 \ y & : 3, 4.6, 5.7, 7, 8.1, 9, 10.2, 10.3 \end{align*} \]

Step 1: Logarithmic Transformation

To use linear regression, take the logarithm of both sides of the power function:

\[ \log(y) = \log(a) + b \log(x) \]

Let \( Y = \log(y) \), \( A = \log(a) \), and \( X = \log(x) \). The equation can be rewritten as:

\[ Y = A + bX \]

Step 2: Calculate \(\log(x)\) and \(\log(y)\)

\[ \begin{align*} x & : 1, 2, 3, 4, 5, 6, 7, 8 \ y & : 3, 4.6, 5.7, 7, 8.1, 9, 10.2, 10.3 \ \log(x) & : 0, 0.301, 0.477, 0.602, 0.699, 0.778, 0.845, 0.903 \ \log(y) & : 0.477, 0.662, 0.755, 0.845, 0.908, 0.954, 1.008, 1.009 \ \end{align*} \]

Step 3: Perform Linear Regression

Next, use the points \((\log(x), \log(y))\):

  1. Calculate the sums needed for the formulas.

    • \( n = 8 \) (number of data points)
    • \( \sum X = 0 + 0.301 + 0.477 + 0.602 + 0.699 + 0.778 + 0.845 + 0.903 \approx 4.605 \)
    • \( \sum Y = 0.477 + 0.662 + 0.755 + 0.845 + 0.908 + 0.954 + 1.008 + 1.009 \approx 6.618 \)
    • \( \sum XY = 00.477 + 0.3010.662 + 0.4770.755 + 0.6020.845 + 0.6990.908 + 0.7780.954 + 0.8451.008 + 0.9031.009 \approx 4.646 \)
    • \( \sum X^2 = 0^2 + 0.301^2 + 0.477^2 + 0.602^2 + 0.699^2 + 0.778^2 + 0.845^2 + 0.903^2 \approx 2.783 \)

  2. Calculate \( b \) and \( A \): \[ b = \frac{n \sum XY - \sum X \sum Y}{n \sum X^2 - (\sum X)^2} \] \[ A = \frac{\sum Y - b \sum X}{n} \]

Step 4: Calculation

  1. Calculate \( b \):

    \[ b = \frac{8(4.646) - (4.605)(6.618)}{8(2.783) - (4.605)^2} \approx \frac{37.168 - 30.487}{22.264 - 21.188} = \frac{6.681}{1.076} \approx 6.199 \]

  2. Calculate \( A \):

    \[ A = \frac{6.618 - 6.199(4.605)}{8} \approx \frac{6.618 - 28.608}{8} \approx -2.124 \]

  3. Finally, back-transform \( A \):

    \[ a = e^A \approx e^{-2.124} \approx 0.12 \]

Final Model

Thus, our power function model is:

\[ y \approx 0.12 x^{6.199} \]

Make sure to check against the data for accuracy. The specific parameters may vary slightly depending on the method of calculation used in regression, but we expect them to be in this general form.

Result:

In the required format:

\[ y \approx 0.120 x^{6.199} \]

Conclusion:

The power function that models the data is \( y = 0.120 x^{6.199} \), rounded to three decimal places.