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.

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12.

The parabola y=ax2+bx+cy = ax^2 + bx + c has vertex (p,p)(p, p) and yy-intercept (0,−p),(0, -p), where p≠0.p \neq 0. What is b?b?
−p-p
00
22
44
pp
Answer: D
Concepts:parabolaquadratic
Small hint:
Write the parabola in vertex form y=a(x−p)2+py = a(x - p)^2 + p
Big hint:
Use the yy-intercept to find aa in terms of p,p, then read off the coefficient of xx
Solution:
The vertex form is y=a(x−p)2+p.y = a(x - p)^2 + p.
At x=0,x = 0, y=ap2+p=−p,y = ap^2 + p = -p, so ap2=−2pap^2 = -2p and a=−2p.a = -\dfrac{2}{p}.
Expanding, y=ax2−2ap x+ap2+p,y = a x^2 - 2ap\, x + ap^2 + p, so b=−2ap=−2(−2p)p=4.b = -2ap = -2\left(-\dfrac{2}{p}\right)p = 4.
Thus, the correct answer is D.
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