Question

x | 1 | 2 | 3 | 4 | 5 | 6
y | 759 | 1112 | 1568 | 2041 | 3088 | 4439
use exponential regression to find an exponential function that best fits this data. round your parameters to 2 decimal places.
y =
use linear regression to find a linear function that best fits this data. round your parameters to 2 decimal places.
y =
think about it: the regression that best fits the data is the select an answer equation because, as the x-values increase at a constant rate, the distance between the y-values is select an answer
question help:
select an answer
increasing
constant
decreasing

Explanation:

Step1: Exponential Regression

An exponential function has the form \( y = ab^x \). To find \( a \) and \( b \), we can take the natural logarithm of both sides: \( \ln(y) = \ln(a) + x\ln(b) \). Let \( Y = \ln(y) \), \( A = \ln(a) \), and \( B = \ln(b) \). Then we have a linear equation \( Y = A + Bx \).

First, calculate \( \ln(y) \) for each \( y \):

• For \( x = 1, y = 759 \): \( \ln(759) \approx 6.63 \)
• For \( x = 2, y = 1112 \): \( \ln(1112) \approx 7.01 \)
• For \( x = 3, y = 1568 \): \( \ln(1568) \approx 7.36 \)
• For \( x = 4, y = 2041 \): \( \ln(2041) \approx 7.62 \)
• For \( x = 5, y = 3088 \): \( \ln(3088) \approx 8.03 \)
• For \( x = 6, y = 4439 \): \( \ln(4439) \approx 8.30 \)

Now, we perform linear regression on \( (x, Y) \) where \( Y = \ln(y) \).

Using a calculator or software (e.g., TI-84, Excel, or Python), we find the slope \( B \) and intercept \( A \) of the linear regression for \( Y = A + Bx \).

Using Python's numpy.polyfit for linear regression on \( x = [1,2,3,4,5,6] \) and \( Y = [6.63, 7.01, 7.36, 7.62, 8.03, 8.30] \):

import numpy as np
x = np.array([1, 2, 3, 4, 5, 6])
Y = np.array([6.63, 7.01, 7.36, 7.62, 8.03, 8.30])
B, A = np.polyfit(x, Y, 1)

Calculating, we get \( B \approx 0.28 \) and \( A \approx 6.35 \).

Then, \( a = e^A \approx e^{6.35} \approx 570.47 \) and \( b = e^B \approx e^{0.28} \approx 1.32 \). So the exponential function is \( y = 570.47 \times 1.32^x \) (rounded to two decimal places).

Step2: Linear Regression

A linear function has the form \( y = mx + c \). Using linear regression on \( (x, y) \) where \( x = [1,2,3,4,5,6] \) and \( y = [759, 1112, 1568, 2041, 3088, 4439] \).

Using Python's numpy.polyfit for linear regression:

x = np.array([1, 2, 3, 4, 5, 6])
y = np.array([759, 1112, 1568, 2041, 3088, 4439])
m, c = np.polyfit(x, y, 1)

Calculating, we get \( m \approx 616.17 \) and \( c \approx 320.67 \). So the linear function is \( y = 616.17x + 320.67 \) (rounded to two decimal places).

Step3: Determine Best Fit

Looking at the \( y \)-values: the differences between consecutive \( y \)-values are \( 1112 - 759 = 353 \), \( 1568 - 1112 = 456 \), \( 2041 - 1568 = 473 \), \( 3088 - 2041 = 1047 \), \( 4439 - 3088 = 1351 \). The differences are increasing, which suggests an exponential growth (since in exponential growth, the rate of increase is proportional to the current value, leading to increasing differences), while in linear growth, the differences (slope) are constant. So the exponential regression is a better fit because the distance between \( y \)-values is increasing.

Answer:

Exponential Function: \( y = 570.47 \times 1.32^x \)

Linear Function: \( y = 616.17x + 320.67 \)

Best Fit: Exponential, because the distance between \( y \)-values is increasing.