2006 AMC 12B Problem 12 — Solution | LIVE by Po-Shen Loh

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2006 AMC 12B Problem 12

Attempt Problem 12 of the 2006 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

The parabola $y = ax^2 + bx + c$ has vertex $(p, p)$ and $y$ -intercept $(0, -p),$ where $p \neq 0.$ What is $b?$

$-p$

$0$

$2$

$4$

$p$

Answer: D

Concepts:parabola quadratic

Difficulty rating: 1530

Solution:

The vertex form is $y = a(x - p)^2 + p.$

At $x = 0,$ $y = ap^2 + p = -p,$ so $ap^2 = -2p$ and $a = -\dfrac{2}{p}.$

Expanding, $y = a x^2 - 2ap\, x + ap^2 + p,$ so $b = -2ap = -2\left(-\dfrac{2}{p}\right)p = 4.$

Thus, the correct answer is D.

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