Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem)
m (→‎Solution)
Line 7: Line 7:
 
<math> \textbf{(D)}\ \text{a non-horizontal line}\qquad\textbf{(E)}\ \text{the graph of a cubic function} </math>
 
<math> \textbf{(D)}\ \text{a non-horizontal line}\qquad\textbf{(E)}\ \text{the graph of a cubic function} </math>
  
−
==Solution==
+
==Solution 1==
  
 
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:
 
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:
Line 17: Line 17:
  
 
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{(D)} \text{ a non-horizontal line }}</math>
 
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{(D)} \text{ a non-horizontal line }}</math>
 +
 +
==Solution 2: less computation==
 +
 +
WLOG let the parabola be <math>y=x^2</math>, and the reflected parabola is thus <math>y=-x^2</math>. We can also assume <cmath>f(x) = (x+5)^2 = x^2 + 10x + 25</cmath> and <cmath>g(x) = -(x-5)^2 = -x^2 + 10x - 25.</cmath> Therefore, <cmath>(f+g)(x) = x^2 + 10x + 25 - x^2 + 10x - 25 = 20x,</cmath> which is a non-horizontal line <math>\Rightarrow \boxed{\textbf{D}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2003|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2003|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:55, 4 January 2019

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 $\Rightarrow \boxed{\textbf{D}}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_19&oldid=100095"