From $p(x)=ax^2+bx-6$, you can immediately find $p(0)$. Compare that value to $p(6)=5$ to decide whether the parabola opens up or down.

Use that the axis is $\frac{r+s}{2}$ and that $rs=\frac{c}{a}$ for $ax^2+bx+c$.

In Desmos, define the unknown coefficients by creating equations from the conditions:

• From $p(2)=p(10)$, enter:

a*2^2 + b*2 - 6 = a*10^2 + b*10 - 6

• From $p(6)=5$, enter:

a*6^2 + b*6 - 6 = 5

Desmos will solve the system for $a$ and $b$ (it may display the solution directly, or you can click the intersection/solution set). Record the resulting values of $a$ and $b$, and note the sign of $b$.

Use Vieta’s formulas with your solved coefficients:

• Enter -b/a to get $r+s$.
• Enter (-6)/a to get $rs$.

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Then compare these to statements I–III. This will determine which choice must be true.

Since $p(2)=p(10)$ and $2\ne 10$, the axis of symmetry is the midpoint of $2$ and $10$:

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If the $x$-intercepts are $r$ and $s$, then the axis of symmetry is the midpoint of the roots:

Because the axis of symmetry is $x=6$, the vertex occurs at $x=6$. We are told $p(6)=5$, so the vertex has $y$-value $5$.

Also,

If $a>0$, the parabola opens upward and the vertex would be a minimum, meaning $p(x)\ge 5$ for all $x$. But $p(0)=-6<5$, which is impossible.

Therefore, $a<0$.

For $p(x)=ax^2+bx-6$, the axis formula gives

Since $a<0$, it follows that $b>0$.

By Vieta’s formulas, the product of the roots is

Because $a<0$, the value $\frac{-6}{a}$ is positive, so $rs>0$.

From the work above:

• $r+s=12$
• $b>0$
• $rs>0$

Therefore, the correct choice is I, II, and III.