Question 6·200 Super-Hard SAT Math Questions·Advanced Math

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p(x)=ax^2+bx-6

In the given quadratic function, $a$ and $b$ are constants and $a\ne 0$ . If $p(2)=p(10)$ and $p(6)=5$ , the graph of $y=p(x)$ has x-intercepts at $(r,0)$ and $(s,0)$ , where $r<s$ . Which choice must be true?

I. $r+s=12$

II. $b>0$

III. $rs>0$

Equal outputs at two inputs reveal a quadratic’s axis of symmetry, the vertical line that mirrors its graph. The roots mirror across that line too, so it gives their sum. A two-condition Desmos regression can then find the coefficient signs. For the root product, divide the constant term by the leading coefficient; the constant’s sign alone isn’t enough.

Hints

Hint 1
A parabola mirrors across a vertical line called its axis of symmetry. If the outputs at $2$ and $10$ match, where is that line? The two $x$ -intercepts mirror across it too.
Hint 2
To use both conditions in Desmos, rewrite equal outputs as $p(2)-p(10)=0$ and pair that with $p(6)=5$ . A list regression can find $a$ and $b$ without solving two equations by hand.
Hint 3
The roots $r$ and $s$ make $p(x)=0$ . In $p(x)=a(x-r)(x-s)$ , the constant term is $ars$ . Compare it with $-6$ , then check the sign of $a$ before deciding whether $rs$ is positive.

Step-by-step

Approach 1: Use symmetry, then fit the coefficients
Step 1
Find the mirror line
The equal outputs $p(2)=p(10)$ put $2$ and $10$ on opposite sides of the same axis of symmetry, the parabola’s vertical mirror line. That line is halfway between them:
$x=\frac{2+10}{2}=6.$
Step 2
Find the root sum
An $x$ -intercept has output $0$ , so the roots $r$ and $s$ are another mirror pair. Their midpoint is $6$ :
$\frac{r+s}{2}=6.$
Multiply by $2$ :
$r+s=12.$
So I is true. One parabola’s mirror line bisects every pair of inputs with equal outputs.
Step 3
Find the coefficient signs
Type $p(x)=ax^2+bx-6$ , then $[p(2)-p(10),p(6)]\sim[0,5]$ . The first list entry makes the given outputs equal; the second uses $p(6)=5$ . The $\sim$ tells Desmos to fit $a$ and $b$ . Under PARAMETERS, it reports $a\approx-0.30556$ and $b\approx3.66667$ . Since $b>0$ , II is true.
Step 4
Check the root product
Each root makes one factor zero, so write $p(x)=a(x-r)(x-s)$ . Expand to see its constant term:
$p(x)=a\bigl(x^2-(r+s)x+rs\bigr).$
Match that constant term to the given $-6$ :
$ars=-6.$
Divide by the nonzero $a$ :
$rs=\frac{-6}{a}>0.$
Desmos found $a<0$ , so III is true too. All three statements hold. Choice D.

Approach 2: Explain the signs from the vertex
Step 1
Determine which way the parabola opens
The axis is $x=6$ , so $p(6)=5$ is the vertex output, the value at the parabola’s turning point. The given formula also gives $p(0)=-6$ . Since the vertex is higher than that point, the parabola opens down, which means $a<0$ .
Step 2
Relate the middle coefficient to the axis
For $ax^2+bx-6$ , the axis is $x=-\frac{b}{2a}$ . Its position at $6$ gives:
$-\frac{b}{2a}=6.$
Multiply by $-2a$ :
$b=-12a.$
Because $a<0$ , this makes $b>0$ , so II is true.
Step 3
Recover the root sum
The root-sum rule for $ax^2+bx+c$ is $r+s=-\frac{b}{a}$ . Substitute $b=-12a$ :
$r+s=-\frac{-12a}{a}=12.$
So I is true without finding either root.
Step 4
Check the root product
The root-product rule gives $rs=\frac{c}{a}$ . Here $c=-6$ and $a<0$ , so:
$rs=\frac{-6}{a}>0.$
III is true as well, so all three statements hold. Choice D.

Question 6·200 Super-Hard SAT Math Questions·Advanced Math

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