SAT Question a45ffacb: Nonlinear Functions answer and explanation with Desmos solution

It's given that $f\left(x\right)=-a^x+b$. Substituting $-a^x+b$ for $f\left(x\right)$ in the equation $y=f\left(x\right)-15$ yields $y=-a^x+b-15$. It's given that the y-intercept of the graph of $y=f\left(x\right)-15$ is $\left(0,-\frac{99}{7}\right)$. Substituting $0$ for $x$ and $-\frac{99}{7}$ for $y$ in the equation $y=-a^x+b-15$ yields $-\frac{99}{7}=-a^0+b-15$, which is equivalent to $-\frac{99}{7}=-1+b-15$, or $-\frac{99}{7}=b-16$. Adding $16$ to both sides of this equation yields $\frac{13}{7}=b$. It's given that the product of $a$ and $b$ is $\frac{65}{7}$, or $ab=\frac{65}{7}$. Substituting $\frac{13}{7}$ for $b$ in this equation yields $\left(a\right)\left(\frac{13}{7}\right)=\frac{65}{7}$. Dividing both sides of this equation by $\frac{13}{7}$ yields $a=5$.