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

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2006 AMC 10A Problem 8

Below is the professionally curated solution for Problem 8 of the 2006 AMC 10A, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:parabola system of equations

Difficulty rating: 1270

8.

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

$2$

$5$

$7$

$10$

$11$

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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