PSAT Math Multiple-Choice Question 462: Answer and Explanation

Question: 462

If a catapult is used to throw a lead ball from ground level, the path of the ball can be modeled by a quadratic equation, y = ax² + bx + c, where x is the horizontal distance that the ball travels and y is the height of the ball. If one of these catapult-launched lead balls travels 150 feet before hitting the ground and reaches a maximum height of 45 feet, which of the following equations represents its path?

• A. y = -0.008x² + 1.2x
• B. y = -0.008x² - 150x
• C. y = 45x² + 150x
• D. y = 125x² + 25x

Correct Answer: A

Explanation:

A

Difficulty: Hard

Getting to the Answer: Begin by considering the shape of the parabola formed by the path of the ball: because the ball starts and ends at ground level (y = 0) and travels a horizontal distance of 150 feet (x = 150), it must have x-intercepts of (0, 0) and (150, 0). In addition, the question tells you that it reaches a maximum height of 45 feet (y = 45) and, because the vertex is halfway between the two x-intercepts, the vertex must be at (75, 45). The vertex is above the x-intercepts, so this is a downward parabola. You can immediately eliminate choices (C) and (D), which have positive a values, making them upward parabolas. To decide between (A) and (B), try plugging in the coordinates of the vertex or an x-intercept to see which equation holds. For example, here are the calculations if you use (150, 0):

(A): 0 = -0.008(150)² + 1.2(150) = -180 + 180 = 0, keep.

(B): 0 = -0.008(150)² - 150(150) = -180 - 22,500 = -22,680 ≠ 0, eliminate.

Thus, (A) is correct.

If you have time, you could also graph each equation in your graphing calculator and find the one that has a maximum value of 45, which would show again that (A) is the only equation for which this is true.