SAT Question 58b4c6f3: Nonlinear Functions answer and explanation with Desmos solution

Choice B is correct. It’s given that function $h$ is defined by $h(x)=a^x+b$, where $a$ and $b$ are positive constants. It's also given that the graph of $y=h(x)$ in the xy-plane passes through the point $(0,10)$. Substituting $0$ for $x$ and $10$ for $h(x)$ in the equation $h(x)=a^x+b$ yields $10=a^0+b$. Since $a^0=1$, this equation can be rewritten as $10=1+b$. Subtracting $1$ from both sides of this equation yields $9=b$. Substituting $9$ for $b$ in the equation $h(x)=a^x+b$ yields $h(x)=a^x+9$. It's also given that the graph of $y=h(x)$ in the xy-plane passes through the point $(2,13)$. Substituting $2$ for $x$ and $13$ for $h(x)$ in the equation $h(x)=a^x+9$ yields $13=a^2+9$. Subtracting $9$ from both sides of this equation yields $4=a^2$. Taking the square root of both sides of this equation yields $2=a$ and $-2=a$. Since $a$ is positive, it follows that $2=a$. Since $9=b$ and $2=a$, it follows that the value of $ab$ is $(2)(9)$, or $18$.