To determine which function has a negative estimated average rate of change over the interval \([0.1, 0.6]\), we can calculate the average rate of change for both the exponential function \(y = 9(13)^x\) and the quadratic function \(y = 9x^2 + 13x\).
The formula for the average rate of change of a function \(f\) over the interval \([a, b]\) is:
\[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \]
Step 1: Calculate for the exponential function
Let \(f(x) = 9(13)^x\):
- Calculate \(f(0.1)\) and \(f(0.6)\).
\[ f(0.1) = 9(13)^{0.1} \approx 9(1.3498) \approx 12.1482 \]
\[ f(0.6) = 9(13)^{0.6} \approx 9(5.0654) \approx 45.5886 \]
Now calculate the average rate of change:
\[ \text{Average Rate of Change} = \frac{f(0.6) - f(0.1)}{0.6 - 0.1} = \frac{45.5886 - 12.1482}{0.5} \approx \frac{33.4404}{0.5} \approx 66.8808 \]
Step 2: Calculate for the quadratic function
Let \(g(x) = 9x^2 + 13x\):
- Calculate \(g(0.1)\) and \(g(0.6)\).
\[ g(0.1) = 9(0.1)^2 + 13(0.1) = 9(0.01) + 1.3 = 0.09 + 1.3 = 1.39 \]
\[ g(0.6) = 9(0.6)^2 + 13(0.6) = 9(0.36) + 7.8 = 3.24 + 7.8 = 11.04 \]
Now calculate the average rate of change:
\[ \text{Average Rate of Change} = \frac{g(0.6) - g(0.1)}{0.6 - 0.1} = \frac{11.04 - 1.39}{0.5} = \frac{9.65}{0.5} = 19.3 \]
Conclusion
- The average rate of change for the exponential function \(y = 9(13)^x\) is approximately \(66.8808\), which is positive.
- The average rate of change for the quadratic function \(y = 9x^2 + 13x\) is \(19.3\), which is also positive.
Since both functions have positive average rates of change over the interval \([0.1, 0.6]\), the answer is:
neither function.
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