Question

find the exponential model of best fit for the points $(-3,5),(1,12),(5,72),(7,137)$. explain how you got your answer. round values to the nearest hundredth.

Explanation:

Step1: Assume the exponential model

Assume the exponential model is \(y = ab^{x}\).

Step2: Substitute the points into the model

For the point \((- 3,5)\), we have \(5=ab^{-3}=\frac{a}{b^{3}}\). For the point \((1,12)\), we have \(12 = ab^{1}=ab\). From \(12 = ab\), we get \(a=\frac{12}{b}\). Substitute \(a=\frac{12}{b}\) into \(\frac{a}{b^{3}} = 5\), then \(\frac{12}{b\times b^{3}}=5\), \(\frac{12}{b^{4}}=5\), \(b^{4}=\frac{12}{5} = 2.4\), \(b=\sqrt[4]{2.4}\approx1.24\).

Step3: Find the value of \(a\)

Since \(a=\frac{12}{b}\), substituting \(b\approx1.24\) gives \(a=\frac{12}{1.24}\approx9.68\).

Step4: Check with other points

For \(x = 5\), \(y=ab^{5}\approx9.68\times(1.24)^{5}\). \((1.24)^{5}=1.24\times1.24\times1.24\times1.24\times1.24\approx2.98\), \(y\approx9.68\times2.98\approx28.85\) (not good). Let's use a better - method.
We can use the formula for the exponential regression. If we have data points \((x_{i},y_{i})\), \(i = 1,\cdots,n\). The general form of the exponential function is \(y=ab^{x}\), taking the natural logarithm of both sides: \(\ln y=\ln a + x\ln b\). Let \(Y = \ln y\), \(A=\ln a\), \(B = \ln b\), then \(Y=A + Bx\).
For the points \((-3,5)\), \(Y_1=\ln5\approx1.61\), \(x_1=-3\); \((1,12)\), \(Y_2=\ln12\approx2.48\), \(x_2 = 1\); \((5,72)\), \(Y_3=\ln72\approx4.28\), \(x_3 = 5\); \((7,137)\), \(Y_4=\ln137\approx4.92\), \(x_4 = 7\).
The formula for \(B=\frac{n\sum_{i = 1}^{n}x_{i}Y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}Y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}\), \(n = 4\), \(\sum_{i=1}^{4}x_{i}=-3 + 1+5 + 7=10\), \(\sum_{i = 1}^{4}Y_{i}=1.61+2.48 + 4.28+4.92=13.29\), \(\sum_{i=1}^{4}x_{i}Y_{i}=(-3)\times1.61+1\times2.48 + 5\times4.28+7\times4.92=-4.83+2.48+21.4+34.44=53.49\), \(\sum_{i = 1}^{4}x_{i}^{2}=(-3)^{2}+1^{2}+5^{2}+7^{2}=9 + 1+25+49=84\).
\(B=\frac{4\times53.49-10\times13.29}{4\times84 - 10^{2}}=\frac{213.96-132.9}{336 - 100}=\frac{81.06}{236}\approx0.34\).
\(A=\frac{\sum_{i = 1}^{n}Y_{i}-B\sum_{i = 1}^{n}x_{i}}{n}=\frac{13.29-0.34\times10}{4}=\frac{13.29 - 3.4}{4}=\frac{9.89}{4}=2.47\).
Since \(B=\ln b\), \(b = e^{B}\approx e^{0.34}\approx1.41\), since \(A=\ln a\), \(a=e^{A}\approx e^{2.47}\approx12\).

So the exponential model is \(y = 12\times(1.41)^{x}\).

Answer:

The exponential model of best - fit is \(y = 12\times(1.41)^{x}\).