For a quadratic function in the form $h(t)=a{t}^2+bt+c$, the maximum occurs at $t=-\frac{b}{2a}$. Here, $a=-16$ and $b=32$, so

Now evaluate the height at $t=1$:

So the ball reaches a maximum height of $21$ feet.

An x-intercept occurs where $y=0$, so the equation must be zero when $x=2$ and when $x=5$. That means the factors must be $(x-2)$ and $(x-5)$, so an equation that could represent the parabola is $y=(x-2)(x-5)$.

To write $h(t)=-16t^2+48t+5$ in vertex form, factor out $-16$ from the quadratic and linear terms:

Complete the square inside the parentheses:

So,

Distribute $-16$:

So the equivalent vertex form is $h(t)=-16{\left(t-\frac{3}{2}\right)}^2+41$.

For a quadratic equation in the form $x^2+kx+18=0$, Vieta’s formulas say that the sum of the solutions is $-k$ and the product of the solutions is $18$. Since the sum is given as $-9$, it follows that $-k=-9$, so $k=9$. No matter what the individual solutions are, their product must still be $18$. Therefore, the statement that must be true is that the two solutions multiply to $18$.

Set up the area equation:

Expand:

Factor:

So $x=-12$ or $x=5$. Since side lengths must be positive, use $x=5$.

Then the side lengths are $7$ and $11$, so the perimeter is