SAT Question 8490cc45: Nonlinear Functions answer and explanation with Desmos solution

Choice A is correct. The y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane occurs at the point on the graph where $x=0$. In other words, when $x=0$, the corresponding value of $f\left(x\right)$ is the y-coordinate of the y-intercept. Substituting $0$ for $x$ in the given equation yields $f\left(0\right)=\left(-8\right){\left(2\right)}^0+22$, which is equivalent to $f\left(0\right)=\left(-8\right)\left(1\right)+22$, or $f\left(0\right)=14$. Thus, when $x=0$, the corresponding value of $f\left(x\right)$ is $14$. Therefore, the y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane is $\left(0,14\right)$.