SAT Question 7902bed0: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. An equation representing the height above ground $h$, in meters, of a softball $t$ seconds after it is launched by a machine from ground level can be written in the form $h=-a{\left(t-b\right)}^2+c$, where $a$, $b$, and $c$ are positive constants. In this equation, $b$ represents the time, in seconds, at which the softball reaches its maximum height of $c$ meters above the ground. It's given that this softball reaches a maximum height of $51.84$ meters above the ground at $1.8$ seconds; therefore, $b=1.8$ and $c=51.84$. Substituting $1.8$ for $b$ and $51.84$ for $c$ in the equation $h=-a{\left(t-b\right)}^2+c$ yields $h=-a{\left(t-1.8\right)}^2+51.84$. It's also given that this softball hits the ground at $3.6$ seconds; therefore, $h=0$ when $t=3.6$. Substituting $0$ for $h$ and $3.6$ for $t$ in the equation $h=-a{\left(t-1.8\right)}^2+51.84$ yields $0=-a{\left(3.6-1.8\right)}^2+51.84$, which is equivalent to $0=-a{\left(1.8\right)}^2+51.84$, or $0=-3.24a+51.84$. Adding $3.24a$ to both sides of this equation yields $3.24a=51.84$. Dividing both sides of this equation by $3.24$ yields $a=16$. Substituting $16$ for $a$ in the equation $h=-a{\left(t-1.8\right)}^2+51.84$ yields $h=-16{\left(t-1.8\right)}^2+51.84$. Therefore, $h=-16{\left(t-1.8\right)}^2+51.84$ represents the height above ground $h$, in meters, of this softball $t$ seconds after it is launched.