To determine the exponential regression equation that best fits the given data points [tex]\((-4, 6.01)\)[/tex], [tex]\((-3, 6.03)\)[/tex], [tex]\((-2, 6.12)\)[/tex], [tex]\((-1, 6.38)\)[/tex], [tex]\((0, 8)\)[/tex], [tex]\((1, 12)\)[/tex], [tex]\((2, 13)\)[/tex], [tex]\((3, 36)\)[/tex], and [tex]\((4, 88)\)[/tex], we need to match these points to an exponential function of the form [tex]\( y = a \cdot b^x \)[/tex].
Given the four options provided:
- Option A: [tex]\(y = 12.11 \cdot 1.36^x\)[/tex]
- Option B: [tex]\(y = 2.49x^2 + 7.29x + 3.57\)[/tex] (this is a quadratic function, not exponential)
- Option C: [tex]\(y = 4.89 \cdot 1.47^x\)[/tex]
- Option D: [tex]\(y = 1.36 \cdot 12.11^x\)[/tex]
First, we can disregard Option B since it is not an exponential function. Therefore, we compare the remaining options, which are exponential functions.
We know that exponential functions take the form [tex]\( y = a \cdot b^x \)[/tex]. By comparing each option and considering the calculated results through careful analysis, we match the coefficients [tex]\( a \)[/tex] and the base [tex]\( b \)[/tex] to the exponential models:
1. Option A: [tex]\(y = 12.11 \cdot 1.36^x\)[/tex]
- Here, [tex]\(a = 12.11\)[/tex] and [tex]\(b = 1.36\)[/tex].
2. Option C: [tex]\(y = 4.89 \cdot 1.47^x\)[/tex]
- Here, [tex]\(a = 4.89\)[/tex] and [tex]\(b = 1.47\)[/tex].
3. Option D: [tex]\(y = 1.36 \cdot 12.11^x\)[/tex]
- This option presents an unconventional form for an exponential equation, where [tex]\( a = 1.36 \)[/tex] and [tex]\( b = 12.11 \)[/tex].
Given our analysis and considering the fit of these parameters to the actual plotted data points, the most appropriate equation that aligns with the observed pattern is:
Option A: [tex]\(y = 12.11 \cdot 1.36^x\)[/tex]
Therefore, the exponential regression equation that best fits the given data is [tex]\(\boxed{y = 12.11 \cdot 1.36^x}\)[/tex].
