2006 AMC 10A Problem 8 — Solution | LIVE by Po-Shen Loh

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2006 AMC 10A Problem 8

Attempt Problem 8 of the 2006 AMC 10A 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 10A solutions, or check the answer key.

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

8.

A parabola with equation $y = x^2 + bx + c$ passes through the points $(2, 3)$ and $(4, 3).$ What is $c?$

$2$

$5$

$7$

$10$

$11$

Answer: E

Concepts:parabola system of equations

Difficulty rating: 1270

Solution:

Substituting the points gives $3 = 4 + 2b + c$ and $3 = 16 + 4b + c.$ Subtracting yields $0 = 12 + 2b,$ so $b = -6.$

Then $c = 3 - 4 - 2(-6) = 11.$

Thus, the correct answer is E.

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Problem 8 in Other Years

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