1. **State the problem:** We have two functions defined for $x \geq 0$: $$f(x) = 232(0.4)^{x+2}$$ $$g(x) = 232(0.4)(0.4)(0.4)^{x-2}$$ We want to find which function's equation displays the maximum value of the function it defines for $x \geq 0$. 2. **Simplify each function:** For $f(x)$: $$f(x) = 232 \times (0.4)^{x+2} = 232 \times (0.4)^2 \times (0.4)^x = 232 \times 0.16 \times (0.4)^x = 37.12 \times (0.4)^x$$ For $g(x)$: $$g(x) = 232 \times 0.4 \times 0.4 \times (0.4)^{x-2} = 232 \times 0.16 \times (0.4)^{x-2} = 37.12 \times (0.4)^{x-2}$$ Rewrite $g(x)$: $$g(x) = 37.12 \times \frac{(0.4)^x}{(0.4)^2} = 37.12 \times (0.4)^x \times \frac{1}{0.16} = 37.12 \times (0.4)^x \times 6.25 = 232 \times (0.4)^x$$ 3. **Analyze the maximum values for $x \geq 0$:** Both functions are of the form $C \times (0.4)^x$ where $C$ is a constant. - For $f(x)$, the constant coefficient is $37.12$. - For $g(x)$, the constant coefficient is $232$. Since $0.4^x$ is a decreasing function for $x \geq 0$ (because $0 < 0.4 < 1$), the maximum value of each function occurs at $x=0$. Calculate maximum values: $$f(0) = 37.12 \times (0.4)^0 = 37.12 \times 1 = 37.12$$ $$g(0) = 232 \times (0.4)^0 = 232 \times 1 = 232$$ 4. **Conclusion:** The maximum value of $f(x)$ is $37.12$ and the maximum value of $g(x)$ is $232$ for $x \geq 0$. Therefore, the equation that displays the maximum value as a constant or coefficient is $g(x) = 232(0.4)(0.4)(0.4)^{x-2}$. 5. **Using Desmos quickly:** - Enter both functions as $f(x) = 232 \times (0.4)^{x+2}$ and $g(x) = 232 \times 0.4 \times 0.4 \times (0.4)^{x-2}$. - Observe the values at $x=0$ to see which function has the higher maximum. This method quickly shows the maximum values without manual calculation.