Because the graph crosses the $x$-axis at $x=-3$, the factor $(x+3)$ has odd multiplicity. Because the graph just touches the $x$-axis at $x=2$, the factor $(x-2)$ has even multiplicity. Since there are no other $x$-intercepts, the least-degree polynomial that fits the description is

Using the point $(0,-12)$,

so $a=-1$. Therefore

which is degree $3$ with leading coefficient $-1$. So the statement that must be true is that $p(x)$ has degree $3$ and leading coefficient $-1$.

Because the arch touches the ground at $x=0$ and $x=12$, the quadratic can be written in factored form as

At the midpoint, $x=6$, and the height is $9$, so

Thus,

Now evaluate the height at $x=3$:

So the height $3$ feet from the left base is $\boxed{6.75}$ feet.

A quadratic with $x$-intercepts at $-2$ and $6$ can be written in factored form as $y=a(x+2)(x-6)$. Since the graph passes through $(0,-12)$, substitute $x=0$ and $y=-12$:

So $a=1$. The quadratic is therefore

Expanding gives

That matches $x^2-4x-12$.

Because the parabola crosses the $x$-axis at $x=1$ and $x=5$, the polynomial can be written as

Since the graph passes through $(3,-8)$, substitute $x=3$ and $p(3)=-8$:

So $a=2$. Then

Therefore, the value of $p(0)$ is $10$.

Because the graph has intercepts at $t=0$ and $t=6$, the function must include factors $t$ and $(t-6)$. So $h(t)$ can be written as $at(t-6)$. Using the point $(2,32)$ gives

so $a=-4$. Therefore,

This function also makes sense for the situation because the parabola opens downward, so the ball rises and then falls back to the ground.