SAT Question 40491607: Nonlinear Functions answer and explanation with Desmos solution

It's given that in the xy-plane, when the graph of the function $f$, where $y=f\left(x\right)$, is shifted up $6$ units, the resulting graph is defined by the function $g$. Therefore, function $g$ can be defined by the equation $g\left(x\right)=f\left(x\right)+6$. It's given that $f\left(x\right)=\left(x-1\right)\left(x+3\right)\left(x-2\right)$. Substituting $\left(x-1\right)\left(x+3\right)\left(x-2\right)$ for $f\left(x\right)$ in the equation $g\left(x\right)=f\left(x\right)+6$ yields $g\left(x\right)=\left(x-1\right)\left(x+3\right)\left(x-2\right)+6$. For the point $\left(4,b\right)$, the value of $x$ is $4$. Substituting $4$ for $x$ in the equation $g\left(x\right)=\left(x-1\right)\left(x+3\right)\left(x-2\right)+6$ yields $g\left(4\right)=\left(4-1\right)\left(4+3\right)\left(4-2\right)+6$, or $g\left(4\right)=48$. It follows that the graph of $y=g\left(x\right)$ crosses through the point $\left(4,48\right)$. Therefore, the value of $b$ is $48$.