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. (4 points)

Explanation:

Step1: Define exponential model form

Exponential model: $y = ab^x$ (take natural log: $\ln y = \ln a + x\ln b$)

Step2: List transformed data

Points $(x,\ln y)$: $(-3,\ln5≈1.6094)$, $(1,\ln12≈2.4849)$, $(5,\ln72≈4.2767)$, $(7,\ln137≈4.9199)$

Step3: Calculate linear regression for transformed data

Let $Y=\ln y$, $m=\ln b$, $c=\ln a$. Compute slope $m$ and intercept $c$:
$\bar{x}=\frac{-3+1+5+7}{4}=2.5$, $\bar{Y}=\frac{1.6094+2.4849+4.2767+4.9199}{4}≈3.3227$
$SS_{xx}=(-3-2.5)^2+(1-2.5)^2+(5-2.5)^2+(7-2.5)^2=30.25+2.25+6.25+20.25=60$
$SS_{xY}=(-3-2.5)(1.6094-3.3227)+(1-2.5)(2.4849-3.3227)+(5-2.5)(4.2767-3.3227)+(7-2.5)(4.9199-3.3227)≈(-5.5)(-1.7133)+(-1.5)(-0.8378)+(2.5)(0.954)+(3.5)(1.5972)≈9.4232+1.2567+2.385+5.5902≈18.6551$
$m=\frac{SS_{xY}}{SS_{xx}}≈\frac{18.6551}{60}≈0.3109$
$c=\bar{Y}-m\bar{x}≈3.3227-0.3109×2.5≈3.3227-0.7773≈2.5454$

Step4: Find a and b

$a=e^c≈e^{2.5454}≈12.80$, $b=e^m≈e^{0.3109}≈1.36$

Answer:

$y≈12.80×1.36^x$