2003 AMC 12A Problems/Problem 19

Revision as of 19:21, 7 August 2011 by Fermat007 (talk | contribs) (Solution)

Problem

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:

\[(f + g)(x) &= ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c &= 20ax + 10b\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)

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 $\boxed{\textbf{(C)} \text{ a horizontal line }}$

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