In the quadratic function above, and are constants with . The function satisfies and . If the graph of has -intercepts at and where , which of the following must be true? I. II. III.

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Super-Hard SAT Math Question 145 | Free SAT Prep | Aniko

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Question 145·200 Super-Hard SAT Math Questions·Advanced Math

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g(x) = ax^2 + bx - 10

In the quadratic function above, $a$ and $b$ are constants with $a \ne 0$ . The function satisfies $g(1) = g(9)$ and $g(5) = 6$ . If the graph of $y = g(x)$ has $x$ -intercepts at $(r, 0)$ and $(s, 0)$ where $r < s$ , which of the following must be true?

I. $r + s = 10$

II. $b > 0$

III. $rs > 0$

When a quadratic gives equal outputs at two different $x$ -values, the axis of symmetry is their midpoint, immediately giving you $r + s$ . To find the direction the parabola opens, compare a known point (like $g(0)$ from the constant term) with the vertex value—if the vertex is higher but points away from it are lower, the parabola opens downward. Then use the axis formula $h = -\frac{b}{2a}$ and the fact that the product of roots equals $\frac{c}{a}$ to test the remaining statements efficiently.

Hints

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Use the equal-output condition

When $g(1) = g(9)$ , the axis of symmetry lies exactly halfway between $x = 1$ and $x = 9$ .

Determine the direction the parabola opens

You know $g(0) = -10$ from the constant term. Compare this with the vertex value to decide whether the vertex is a maximum or minimum.

Analyze the sign of the product of roots

Once you know the vertex location and direction, consider where the parabola crosses the $x$ -axis relative to the $y$ -axis.

Desmos Guide

Set up the system

In Desmos, define two equations from the conditions:

25a + 5b - 10 = 6 (from $g(5) = 6$ )

b = -10a (from axis of symmetry at $x = 5$ )

Solve for $a$ and $b$

Substitute to get 25a + 5(-10a) - 10 = 6, which gives 25a - 50a = 16, so a = -16/25. Then b = -10(-16/25) = 32/5.

Graph and verify

Enter g(x) = (-16/25)x^2 + (32/5)x - 10. Click the $x$ -intercepts to find $r$ and $s$ , and verify that $r + s = 10$ , $b > 0$ , and $rs > 0$ .

Step-by-step Explanation

Find the axis of symmetry and sum of roots

Since $g(1) = g(9)$ and $1 \ne 9$ , the axis of symmetry is the midpoint:

\frac{1 + 9}{2} = 5

The roots of a parabola are symmetric about the axis of symmetry:

\frac{r + s}{2} = 5 \implies r + s = 10

Statement I is true.

Determine the sign of $a$

The vertex occurs at $x = 5$ , and we are given $g(5) = 6$ .

From the function, $g(0) = -10$ .

If $a > 0$ , the parabola opens upward and the vertex is a minimum, so all other points would be at or above the vertex value of $6$ . But $g(0) = -10 < 6$ , which contradicts this.

Therefore, $a < 0$ (the parabola opens downward).

Determine the sign of $b$

Using the axis of symmetry formula:

5 = -\frac{b}{2a} \implies b = -10a

Since $a < 0$ , we have $b = -10a > 0$ .

Statement II is true.

Determine the sign of $rs$

For a quadratic $ax^2 + bx + c$ , the product of the roots equals $\frac{c}{a}$ :

rs = \frac{c}{a} = \frac{-10}{a}

Since $a < 0$ , we have $rs = \frac{-10}{a} > 0$ .

Statement III is true.

All three statements are true, so the answer is I, II, and III.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation