SAT Question 2b1a27cd: Nonlinear Functions answer and explanation with Desmos solution

The quadratic function $g$ gives the estimated depth of the seal, $g\left(t\right)$, in meters, $t$ minutes after the seal enters the water. It's given that function $g$ estimates that the seal reached its maximum depth of $302.4$ meters $6$ minutes after it entered the water. Therefore, function $g$ can be expressed in vertex form as $g\left(t\right)=a{\left(t-6\right)}^2+302.4$, where $a$ is a constant. Since it's also given that the seal reached the surface of the water after $12$ minutes, $g\left(12\right)=0$. Substituting $12$ for $t$ and $0$ for $g\left(t\right)$ in $g\left(t\right)=a{\left(t-6\right)}^2+302.4$ yields $0=a{\left(12-6\right)}^2+302.4$, or $36a=-302.4$. Dividing both sides of this equation by $36$ gives $a=-8.4$. Substituting $-8.4$ for $a$ in $g\left(t\right)=a{\left(t-6\right)}^2+302.4$ gives $g\left(t\right)=-8.4{\left(t-6\right)}^2+302.4$. Substituting $10$ for $t$ in $g\left(t\right)$ gives $g\left(10\right)=-8.4{\left(10-6\right)}^2+302.4$, which is equivalent to $g\left(10\right)=-8.4{\left(4\right)}^2+302.4$, or $g\left(10\right)=168$. Therefore, the estimated depth, to the nearest meter, of the seal $10$ minutes after it entered the water was $168$ meters.