Since

substitute $x=-1$. Then the right-hand side is

So

which means

Therefore, this statement must be true.

A graph that crosses the $x$-axis at $x=-3$ has a factor of $(x+3)$ with odd multiplicity. A graph that just touches the $x$-axis at $x=2$ has a factor of $(x-2)$ with even multiplicity, so $(x-2{)}^2$ fits. That leaves options with the form $k(x+3)(x-2{)}^2$, where $k$ is positive or negative. Use the point $(0,-12)$:

Since $p(0)=-12$, we get $12k=-12$, so $k=-1$. Therefore,

Factor the numerator: $x^2-9=(x-3)(x+3)$. Then

For $x\ne 3$, the factor $x-3$ cancels, leaving $x+3$. So the equivalent expression is $x+3$.

Because the graph crosses the x-axis at $x=-1$, the factor $(x+1)$ must have odd multiplicity. Because the graph touches the x-axis at $x=4$ and turns around, the factor $(x-4)$ must have even multiplicity, so a possible form is

Using the point $(0,-8)$ gives

so

Therefore,

From the table, the outputs are the same for opposite values of $x$: $f(-2)=f(2)$ and $f(-1)=f(1)$. That suggests the graph is symmetric about the $y$-axis, so an equation involving only $x^2$ is reasonable. Also, the table shows $f(0)=-1$, so the constant term must be $-1$. Among the choices, $f(x)=x^2-1$ fits that value. Checking the zeros: $(-1{)}^2-1=0$ and $1^2-1=0$, which matches the table. Therefore, the equation is $f(x)=x^2-1$.