AMC10 2006 A

AMC10 2006 A · Q8

AMC10 2006 A · Q8. It mainly tests Systems of equations, Quadratic equations.

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

抛物线方程 $y = x^2 + bx + c$ 经过点 (2,3) 和 (4,3)。$c$ 是多少？

(A) 2 2

(B) 5 5

(D) 10 10

(E) 11 11

Answer

Correct choice: (E)

正确答案：（E）

Solution

(E) Substitute $(2,3)$ and $(4,3)$ into the equation to give $3 = 4 + 2b + c$ and $3 = 16 + 4b + c$. Subtracting corresponding terms in these equations gives $0 = 12 + 2b$. So $b = -6$ and $c = 3 - 4 - 2(-6) = 11$. OR The parabola is symmetric about the vertical line through its vertex, and the points $(2,3)$ and $(4,3)$ have the same $y$-coordinate. The vertex has $x$-coordinate $(2 + 4)/2 = 3$, so the equation has the form $y = (x - 3)^2 + k$ for some constant $k$. Since $y = 3$ when $x = 4$, we have $3 = 1^2 + k$ and $k = 2$. Consequently the constant term $c$ is $(-3)^2 + k = 9 + 2 = 11.$

（E）将$(2,3)$和$(4,3)$代入方程得到 $3 = 4 + 2b + c$以及$3 = 16 + 4b + c$。将这两个方程对应项相减得$0 = 12 + 2b$。因此 $b = -6$，且$c = 3 - 4 - 2(-6) = 11$。或者抛物线关于过其顶点的竖直直线对称，而点$(2,3)$与$(4,3)$具有相同的$y$坐标。顶点的$x$坐标为$(2 + 4)/2 = 3$，所以方程可写成 $y = (x - 3)^2 + k$ 其中$k$为常数。由于当$x = 4$时$y = 3$，有$3 = 1^2 + k$，所以$k = 2$。因此常数项$c$为 $(-3)^2 + k = 9 + 2 = 11$。

Topics