Nonlinear functions SAT question

The correct answer is $3.44$. It’s given that $f(0)=86$ and that for each increase in $x$ by $1$, the value of $f(x)$ decreases by $80\%$. Because the output of the function decreases by a constant percentage for each $1$-unit increase in the value of $x$, this relationship can be represented by an exponential function of the form $f(x)=a{(b)}^x$, where $a$ represents the initial value of the function and $b$ represents the rate of decay,
expressed as a decimal. Because $f(0)=86$, the value of $a$ must be $86$. Because the value of $f(x)$ decreases by $80\%$ for each $1$-unit increase in $x$, the value of $b$ must be $(1-0.80)$, or $0.2$. Therefore, the function $f$ can be defined by $f(x)=86{(0.2)}^x$. Substituting $2$ for $x$ in this function yields $f(2)=86{(0.2)}^2$, which is equivalent to $f(2)=86(0.04)$, or $f(2)=3.44$. Either $3.44$ or $86/25$ may be entered as the correct answer.

Alternate approach: It’s given that $f(0)=86$ and that for each increase in $x$ by $1$, the value of $f(x)$ decreases by $80\%$. Therefore, when $x=1$, the value of $f(x)$ is $(100-80)\%$, or $20\%$, of $86$, which can be expressed as $(0.20)(86)$. Since $(0.20)(86)=17.2$, the value of $f(1)$ is $17.2$. Similarly, when $x=2$, the value of $f(x)$ is $20\%$ of $17.2$, which can be expressed as $(0.20)(17.2)$. Since $(0.20)(17.2)=3.44$, the value of $f(2)$ is $3.44$. Either $3.44$ or $86/25$ may be entered as the correct answer.