Algebra 1 Chapter 6 Test Review 12) For the function \( f(x)=2(5)^{x} \), describe the following: Domain: \( \qquad \) Range: \( \qquad \) Asymptote: \( \qquad \) Growth or Decay: \( \qquad \) 13) Ms. Timarky deposits \( \$ 300 \) in an account that earns simple interest at an annual rate of \( 2 \% \). Ms. Waugh deposits \( \$ 300 \) in an account that earns \( 2 \% \) interest and is compounded annually. After 5 years, who has more money in their account and why? 14) An exponential function can be modeled by the function \( P=2(1.45)^{x} \). Which of the following statements is true? Select all that apply. A. \( \square \) This function models exponential growth. B. \( \square \) This function models exponential decay. C. \( \square \) The initial amount is 0.45 . D. \( \square \) The initial amount is 2 . E. \( \square \) The rate is \( 1.45 \% \). F. \( \square \) The rate is \( 45 \% \).

For the function \( f(x)=2(5)^{x} \): The domain is all real numbers, as you can input any value for \( x \) without restriction. The range is all positive real numbers, since the function never crosses the x-axis and only produces values greater than zero. The asymptote is the horizontal line \( y=0 \), which the graph approaches but never touches as \( x \) decreases. This function exhibits growth since the base (5) is greater than 1, indicating that as \( x \) increases, \( f(x) \) will also increase. When comparing Ms. Timarky's and Ms. Waugh's investments after 5 years, Ms. Waugh will have more money in her account. Simple interest is calculated linearly, yielding \( \$ 300 + (\$ 300 \times 0.02 \times 5) = \$ 330 \). In contrast, Ms. Waugh’s compounded interest grows exponentially, calculated as \( P = 300(1 + 0.02)^5 \). After 5 years, her account would show approximately \( \$ 330.60 \), outpacing Ms. Timarky’s total. So, compounding works wonders! For the function \( P=2(1.45)^{x} \): The correct choices are A and D. This function models exponential growth because the base (1.45) is greater than 1, indicating the amount increases as \( x \) increases. The initial amount is indeed 2, which represents the value of \( P \) when \( x=0 \). So, while it may look tempting to dive into the percentages thinking of decay, this function is clearly growing!