SAT Question 1853bb35: Nonlinear Functions answer and explanation with Desmos solution

Choice C is correct. Since the value of $q\left(x\right)$ decreases by a fixed percentage, $45\%$, for every increase in the value of $x$ by $1$, the function $q$ is a decreasing exponential function. A decreasing exponential function can be written in the form $q\left(x\right)=a{\left(1-\frac{p}{100}\right)}^x$, where $a$ is the value of $q\left(0\right)$ and the value of $q\left(x\right)$ decreases by $p\%$ for every increase in the value of $x$ by $1$. If $q\left(0\right)=14$, then $a=14$. Since the value of $q\left(x\right)$ decreases by $45\%$ for every increase in the value of $x$ by $1$, $p=45$. Substituting $14$ for $a$ and $45$ for $p$ in the equation $q\left(x\right)=a{\left(1-\frac{p}{100}\right)}^x$ yields $q\left(x\right)=14{\left(1-\frac{45}{100}\right)}^x$, which is equivalent to $q\left(x\right)=14{\left(1-0.45\right)}^x$, or $q\left(x\right)=14{\left(0.55\right)}^x$.