Question

Given an exponential function for compounding interest, A(x) = P(1.04)x, what is the rate of change? (1 point) A. 4% B. 1.04% C. 104% D. 0.04%

Answer 1

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year} \end{array}\\ t=years \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf A(x)=P(1.04)^x\implies A(x)=P(1+0.04)^x\implies A(x)=P\left(1+\frac{0.04}{1} \right)^x \\\\[-0.35em] ~\dotfill\\\\ r=0.04\implies \stackrel{\textit{to percentage}}{r\% = 0.04\cdot 100}\implies r=\stackrel{\%}{4}$