Difference between revisions of "2003 AMC 12A Problems/Problem 19"

(Created page with '==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 tha…')
 
(Solution)
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==Solution==
 
==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:
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If we take the parabola <math>ax^2 + bx + c</math> and reflect it over the x - axis, we have the parabola <math>-ax^2 - bx - c</math>. Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then:
 
   
 
   
f(x) = a(x+5)^2 + b(x+5) + c = ax^2 + 10ax + 25a + bx +5b +c = ax^2 + (10a+b)x + 25a + 5b + c.
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<cmath> \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*} </cmath>
g(x) = -a(x-5)^2 - b(x-5) - c = -ax^2 + 10ax -bx - 25a + 5b - c
 
  
Adding them up:
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Adding them up produces: <cmath> (f + g)(x) &= ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c &= 20ax + 10b </cmath>
  
(f+g)(x) = 20ax + 10b which is a line with slope 20a. We know this is not horizontal, as a is not equal to 0. This is true because if a = 0, then the graph of ax^2 + bx +c would not be a parabola as stated in the problem.
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This is a line with slope <math>20a</math>. Since <math>a</math> cannot be <math>0</math> (because <math>ax^2 + bx + c</math> would be a line) we end up with <math>\boxed{\textbf{(C)} \text{ a horizontal line }}</math>

Revision as of 19:21, 7 August 2011

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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