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2003 AMC 12A Problems/Problem 19

Revision as of 21:55, 4 January 2019 by Mp8148 (talk | contribs) (→‎Solution 2: less computation)
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Problem

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 $y=(f+g)(x)?$

$\textbf{(A)}\ \text{a parabola tangent to the }x\text{-axis}$ $\textbf{(B)}\ \text{a parabola not tangent to the }x\text{-axis}\qquad\textbf{(C)}\ \text{a horizontal line}$ $\textbf{(D)}\ \text{a non-horizontal line}\qquad\textbf{(E)}\ \text{the graph of a cubic function}$

Solution 1

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 $\boxed{\textbf{(D)} \text{ a non-horizontal line }}$

Solution 2: less computation

WLOG let the parabola be $y=x^2$, and the reflected parabola is thus $y=-x^2$. We can also assume \[f(x) = (x+5)^2 = x^2 + 10x + 25\] and \[g(x) = -(x-5)^2 = -x^2 + 10x - 25.\] Therefore, \[(f+g)(x) = x^2 + 10x + 25 - x^2 + 10x - 25 = 20x,\] which is a non-horizontal line $\Rightarrow \boxed{\textbf{D}}$.

-MP8148

See Also

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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