SAT Question d41cf4d3: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. It's given that $f\left(24\right)<0$. Substituting $24$ for $f\left(x\right)$ in the equation $f\left(x\right)=a\sqrt{x+b}$ yields $f\left(24\right)=a\sqrt{24+b}$. Therefore, $a\sqrt{24+b}<0$. Since $\sqrt{24+b}$ can't be negative, it follows that $a<0$. It's also given that the graph of $y=f\left(x\right)$ passes through the point $\left(-24,0\right)$. It follows that when $x=-24$, $f\left(x\right)=0$. Substituting $-24$ for $x$ and $0$ for $f\left(x\right)$ in the equation $f\left(x\right)=a\sqrt{x+b}$ yields $0=a\sqrt{-24+b}$. By the zero product property, either $a=0$ or $\sqrt{-24+b}=0$. Since $a<0$, it follows that $\sqrt{24+b}=0$. Squaring both sides of this equation yields $-24+b=0$. Adding $24$ to both sides of this equation yields $b=24$. Since $a<0$ and $b$ is $24$, it follows that $a<b$ must be true.