SAT Question a58232b7: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. A function defined by an equation in the form $f\left(x\right)=a{\left(b\right)}^{x+h}$, where $a$, $b$, and $h$ are positive constants and $x\ge 0$, has a minimum value of $f(0)$. It's given that function $g$ is defined by $g\left(x\right)=18\left(1.16\right){\left(1.4\right)}^{x+2}$, which is equivalent to $g\left(x\right)=20.88{\left(1.4\right)}^{x+2}$. Substituting $0$ for $x$ in this equation yields $g\left(0\right)=20.88(1.4{)}^{0+2}$, or $g(0)=40.9248$. Therefore, the minimum value of $g(x)$ is $40.9248$, so $g\left(x\right)=18\left(1.16\right){\left(1.4\right)}^{x+2}$ doesn't display its minimum value as a constant or coefficient. It's also given that function $h$ is defined by $h\left(x\right)=18{\left(1.4\right)}^{x+4}$. Substituting $0$ for $x$ in this equation yields $h\left(0\right)=18{\left(1.4\right)}^{0+4}$, or $h(0)=69.1488$. Therefore, the minimum value of $h(x)$ is $69.1488$, so $h\left(x\right)=18{\left(1.4\right)}^{x+4}$ doesn't display its minimum value as a constant or coefficient. Therefore, neither I nor II displays, as a constant or coefficient, the minimum value of the function it defines, where $x\ge 0$.