AMC12 2003 A

AMC12 2003 A · Q19

AMC12 2003 A · Q19. It mainly tests Graphs (coordinate plane), Transformations.

A parabola with equation $y=ax^2+bx+c$ is reflected about the $x$-axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of $y=f(x)$ and $y=g(x)$, respectively. Which of the following describes the graph of $y=(f+g)(x)?$

将方程为 $y=ax^2+bx+c$ 的抛物线关于 $x$ 轴反射。然后将该抛物线与其反射图像分别向相反方向水平平移 $5$ 个单位，得到 $y=f(x)$ 与 $y=g(x)$ 的图像。下列哪一项描述了 $y=(f+g)(x)$ 的图像？

(A) a parabola tangent to the x-axis a parabola tangent to the x-axis

(B) a parabola not tangent to the x-axis a parabola not tangent to the x-axis

(D) a non-horizontal line a non-horizontal line

(E) the graph of a cubic function the graph of a cubic function

Answer

Correct choice: (D)

正确答案：（D）

Solution

If we take the parabola $ax^2 + bx + c$ and reflect it over the x - axis, we have the parabola $-ax^2 - bx - c$. Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then: \begin{align*} f(x) = a(x+5)^2 + b(x+5) + c = ax^2 + (10a+b)x + 25a + 5b + c \\ g(x) = -a(x-5)^2 - b(x-5) - c = -ax^2 + 10ax -bx - 25a + 5b - c \end{align*} Adding them up produces: \begin{align*} (f + g)(x) = ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c = 20ax + 10b \end{align*} This is a line with slope $20a$. Since $a$ cannot be $0$ (because $ax^2 + bx + c$ would be a line) we end up with $\boxed{\textbf{(D)} \text{ a non-horizontal line }}$

将抛物线 $ax^2 + bx + c$ 关于 $x$ 轴反射，得到抛物线 $-ax^2 - bx - c$。不失一般性，设原抛物线向左平移 $5$ 个单位，而反射后的抛物线向右平移 $5$ 个单位。则： \begin{align*} f(x) = a(x+5)^2 + b(x+5) + c = ax^2 + (10a+b)x + 25a + 5b + c \\ g(x) = -a(x-5)^2 - b(x-5) - c = -ax^2 + 10ax -bx - 25a + 5b - c \end{align*} 将它们相加得到： \begin{align*} (f + g)(x) = ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c = 20ax + 10b \end{align*} 这是一条斜率为 $20a$ 的直线。由于 $a$ 不能为 $0$（否则 $ax^2 + bx + c$ 将是一条直线），因此图像为 $\boxed{\textbf{(D)} \text{一条非水平直线}}$。

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