Nonlinear functions SAT question

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

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

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

Choice D is incorrect. This could be the y-intercept for $f(x)=(-8){(2)}^x$, not $f(x)=(-8){(2)}^x+22$.