AMC12 2001 A

AMC12 2001 A · Q13

AMC12 2001 A · Q13. It mainly tests Functions basics, Transformations.

The parabola with equation $p(x) = ax^2+bx+c$ and vertex $(h,k)$ is reflected about the line $y=k$. This results in the parabola with equation $q(x) = dx^2+ex+f$. Which of the following equals $a+b+c+d+e+f$?

方程为 $p(x) = ax^2+bx+c$、顶点为 $(h,k)$ 的抛物线关于直线 $y=k$ 反射，得到方程为 $q(x) = dx^2+ex+f$ 的抛物线。以下哪一项等于 $a+b+c+d+e+f$？

(A) $2b$ $2b$

(B) $2c$ $2c$

(D) $2h$ $2h$

(E) $2k$ $2k$

Answer

Correct choice: (E)

正确答案：（E）

Solution

We write $p(x)$ as $a(x-h)^2+k$ (this is possible for any parabola). Then the reflection of $p(x)$ is $q(x) = -a(x-h)^2+k$. Then we find $p(x) + q(x) = 2k$. Since $p(1) = a+b+c$ and $q(1) = d+e+f$, we have $a+b+c+d+e+f = 2k$, so the answer is $\fbox{E}$.

将 $p(x)$ 写成 $a(x-h)^2+k$（任意抛物线都可以这样表示）。则 $p(x)$ 的反射为 $q(x) = -a(x-h)^2+k$。于是 $p(x) + q(x) = 2k$。由于 $p(1) = a+b+c$ 且 $q(1) = d+e+f$，所以 $a+b+c+d+e+f = 2k$，因此答案是 $\fbox{E}$。

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