From the table, $f(1)=0$ and $f(3)=0$, so the quadratic has zeros at $x=1$ and $x=3$. That means it must have the form $f(x)=a(x-1)(x-3)$. Use another point from the table, such as $f(2)=-1$:

Since $f(2)=-1$, it follows that $a=1$. So

This also matches the table values, such as $f(0)=3$ and $f(-1)=8$.

Let the integer solutions be $n$ and $n+1$. Since they are roots of $x^2+bx+12=0$, their product is $12$. So

The pair of consecutive integers that multiplies to $12$ is $3$ and $4$. Therefore, the two solutions are $3$ and $4$. For a quadratic $x^2+bx+12=0$, the sum of the roots is $-b$. Since $3+4=7$, we have $-b=7$, so $b=-7$. This must be true.

Because the coefficient of $x^2$ is negative, the parabola opens downward, so the greatest profit occurs at the vertex. For

the $x$-coordinate of the vertex is

So the profit is greatest when the ticket price is \20$.$

For the equation

the solutions $x$ and $y$ satisfy $x+y=8$ and $xy=12$. To find $x^2+y^2$, use the identity

Substituting gives

So the correct answer is $40$.

Factor the numerator as a difference of squares:

Then, for $x\ne 3$,

The restriction $x\ne 3$ matters because the original expression is undefined at $x=3$, even though the simplified form is $x+3$.