Nonlinear functions SAT question

Choice C is correct. Since the value of $q(x)$ 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 $q(x)=a{(1-\frac{p}{100})}^x$, where $a$ is the value of $q(0)$ and the value of $q(x)$ decreases by $p\%$ for every increase in the value of $x$ by $1$. If $q(0)=14$, then $a=14$. Since the value of $q(x)$ 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 $q(x)=a{(1-\frac{p}{100})}^x$ yields $q(x)=14{(1-\frac{45}{100})}^x$, which is equivalent to $q(x)=14{(1-0.45)}^x$, or $q(x)=14{(0.55)}^x$.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect. For this function, the value of $q(x)$ increases, rather than decreases, by $45\%$ for every increase in the value of $x$ by $1$.