Multiple choice

The quadratic function f(x) = ax2 + bx + c, where a, b, and c are constants. The graph of y = f(x) in the xy-plane passes through the points (−2, 0) and (12, 0). If a is an integer greater than 1, what is the greatest possible value of 5b − 2a?

  1. −200

  2. −156

  3. −130

  4. −104
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The roots are -2 and 12, so f(x) = a(x + 2)(x - 12) = a(x^2 - 10x - 24) = ax^2 - 10ax - 24a. Here b = -10a. We want to maximize 5b - 2a = 5(-10a) - 2a = -52a. Since a is an integer greater than 1, the maximum value occurs at the smallest possible a, which is a=2. Thus, -52 * 2 = -104.

AI explanation

Since the graph passes through the points (negative 2, 0) and (12, 0), the equation can be written as y equals a times the quantity x plus 2 times the quantity x minus 12. Expanding this gives y equals a x squared minus 10 a x minus 24 a, meaning the constants are b equals negative 10 a and c equals negative 24 a. To maximize 5b minus 2a, which simplifies to negative 52 a, we must use the smallest possible integer greater than 1, so a equals 2. Multiplying negative 52 by 2 yields negative 104.

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