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).

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

The two points have the same

y

-value, so they are symmetric about the vertex

The axis of symmetry is

x = 3,

giving

b = -6

Substituting the points gives

3 = 4 + 2b + c

and

3 = 16 + 4b + c.

Subtracting yields

0 = 12 + 2b,

b = -6.

Then

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

Thus, the correct answer is E.

Problem 8 in other years