SAT Question 512bd5b4: Nonlinear Functions answer and explanation with Desmos solution

For a function defined by an equation of the form $f\left(x\right)=a{\left(1+\frac{r}{100}\right)}^{\frac{x}{k}}$, where $a$, $r$, and $k$ are positive constants, the value of $f\left(x\right)$ increases by $r\%$ for every increase of $x$ by $k$. The given equation can be rewritten as $f\left(x\right)=27{\left(1.20\right)}^{\frac{2x}{6}}$, or $f\left(x\right)=27{\left({\left(1.2\right)}^2\right)}^{\frac{x}{6}}$, which is equivalent to $f\left(x\right)=27{\left(1.44\right)}^{\frac{x}{6}}$, or $f\left(x\right)=27{\left(1+\frac{44}{100}\right)}^{\frac{x}{6}}$. This equation is of the form $f\left(x\right)=a{\left(1+\frac{r}{100}\right)}^{\frac{x}{k}}$, where $a=27$, $r=44$, and $k=6$. It follows that the value of $f\left(x\right)$ increases by $44\%$ for every increase of $x$ by $6$. Therefore, the value of $p$ is $44$.

Alternate approach: It's given that $f\left(x\right)=27{\left(1.20\right)}^{\frac{x}{3}}$. An increase in the value of $x$ by $6$ can be represented by the expression $x+6$. The value of $f\left(x+6\right)$ can be found by substituting $x+6$ for $x$ in the equation $f\left(x\right)=27{\left(1.20\right)}^{\frac{x}{3}}$, which yields $f\left(x+6\right)=27{\left(1.20\right)}^{\frac{x+6}{3}}$. By the product property of exponents, this equation is equivalent to $f\left(x+6\right)=27{\left(1.20\right)}^{\frac{x}{3}}\cdot {\left(1.20\right)}^{\frac{6}{3}}$, or $f\left(x+6\right)=27{\left(1.20\right)}^{\frac{x}{3}}\cdot {\left(1.20\right)}^2$, which can be rewritten as $f\left(x+6\right)=27{\left(1.20\right)}^{\frac{x}{3}}\cdot \left(1.44\right)$. Since $f\left(x\right)=27{\left(1.20\right)}^{\frac{x}{3}}$, it follows that $f(x+6)=f(x)\cdot (1.44)$. Therefore, for any value of $x$, the value of $f(x+6)$ is $1.44$ times, or $44\%$ greater than, the value of $f\left(x\right)$. It's given that the value of $f\left(x\right)$ increases by $p\%$ for every increase in the value of $x$ by $6$. It follows that the value of $p$ is $44$.