SAT Question de39858a: Nonlinear Functions answer and explanation with Desmos solution

Choice C is correct. It's given that the function $h$ is defined by $h\left(x\right)=a^x+b$ and that the graph of $y=h(x)$ in the xy-plane passes through the points $\left(0,10\right)$ and $\left(-2,\frac{325}{36}\right)$. Substituting $0$ for $x$ and $10$ for $h\left(x\right)$ in the equation $h\left(x\right)=a^x+b$ yields $10=a^0+b$, or $10=1+b$. Subtracting $1$ from both sides of this equation yields $9=b$. Substituting $-2$ for $x$ and $\frac{325}{36}$ for $h\left(x\right)$ in the equation $h(x)=a^x+9$ yields $\frac{325}{36}=a^{-2}+9$. Subtracting $9$ from both sides of this equation yields $\frac{1}{36}=a^{-2}$, which can be rewritten as $a^2=36$. Taking the square root of both sides of this equation yields $a=6$ and $a=-6$, but because it's given that $a$ is a positive constant, $a$ must equal $6$. Because the value of $a$ is $6$ and the value of $b$ is $9$, the value of $ab$ is $(6)(9)$, or $54$.