SAT Question 0de8359f: Nonlinear Functions answer and explanation with Desmos solution

Choice B is correct. It's given that $s\left(t\right)=r\left(t\right)+1$. Since $r\left(t\right)=63t-2t^2$, it follows that $s\left(t\right)=63t-2t^2+1$, or $s\left(t\right)=-2t^2+63t+1$. For a quadratic function $s$ defined by an equation of the form $s\left(t\right)=a{\left(t-h\right)}^2+k$, where $a$, $h$, and $k$ are constants and $a$ is negative, the value of $s(t)$ reaches its maximum, $k$, when the value of $t$ is $h$. The equation $s\left(t\right)=-2t^2+63t+1$ can be rewritten in the form $s\left(t\right)=a(t-h{)}^2+k$ by completing the square. The equation $s(t)=-2t^2+63t+1$ is equivalent to $s\left(t\right)=-2\left(t^2-\frac{63}{2}t\right)+1$, or $s\left(t\right)=-2\left(t^2-\frac{63}{2}t+{\left(\frac{63}{4}\right)}^2-{\left(\frac{63}{4}\right)}^2\right)+1$. This equation can be rewritten as $s\left(t\right)=-2\left({\left(t-\frac{63}{4}\right)}^2-{\left(\frac{63}{4}\right)}^2\right)+1$, or $s\left(t\right)=-2{\left(t-\frac{63}{4}\right)}^2+1+2{\left(\frac{63}{4}\right)}^2$. This equation is in the form $s\left(t\right)=a{\left(t-h\right)}^2+k$, where $a=-2$, $h=\frac{63}{4}$, and $k=1+2{\left(\frac{63}{4}\right)}^2$. Therefore, the maximum value of $s\left(t\right)$ is $1+2{\left(\frac{63}{4}\right)}^2$.