let's first find the places where the ball hits the ground, that way we know how many seconds it took, let's set f(t) = 0 like before, check the picture below.
[tex] \bf \stackrel{f(t)}{0}=-16t^2+44t+12\implies 0=-4t^2+11t+3\\\\\\4t^2-11t-3=0\implies (4t+1)(t-3)=0\implies t=\begin{cases}3\\-\frac{1}{4}\end{cases} [/tex]
so we know the graph goes from -1/4 to 3, so the ball hits the ground on its way back down at x = 3.
now, let's find the maximum, we only need the x-coordinate from the vertex,
[tex] \bf \textit{vertex of a vertical parabola, using coefficients}\\\\f(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+44}t\stackrel{\stackrel{c}{\downarrow }}{+12}\qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)\\\\\\\left(-\cfrac{44}{2(-16)}~~,\qquad \right)\implies \left(\cfrac{11}{8}~~,\qquad \right) [/tex]
so, the ball reaches at maximum when x = 11/8, or 1.375, and hits the ground when x = 3.
so the domain, namely the values for x are, x ⩾ 1.375 and x ⩽ 3, or we can write it as a triplet of 1.375 ⩽ x ⩽ 3.
