SAT Question f01108a8: Nonlinear Functions answer and explanation with Desmos solution

Choice B is correct. It’s given that $y=9{\left(\frac{a}{7}\right)}^{x+c}-b$, where $a$, $b$, and $c$ are positive constants such that $a>7$ and $b>c$. The graph of any equation in the xy-plane crosses the x-axis when there is a point on the graph for which the value of $y$ is equal to $0$, and the value of $y$ changes sign when the graph passes through that point. Since $a>7$, the base $\frac{a}{7}$ is greater than $1$, so the expression $9{\left(\frac{a}{7}\right)}^{x+c}-b$ increases as $x$ increases. Since $b$ is positive, there is a point on the graph of $y=9{\left(\frac{a}{7}\right)}^{x+c}-b$ for which the value of $y$ is equal to $0$. When the graph passes through this point, the value of $y$ changes sign from negative to positive. Therefore, the graph of the given equation crosses the x-axis exactly one time.