Difference between revisions of "2003 AMC 12A Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | A parabola with equation <math>y=ax^2+bx+c</math> is reflected about the <math>x</math>-axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of <math>y=f(x)</math> and <math>y=g(x)</math>, respectively. Which of the following describes the graph of <math>y=(f+g)x</math> | + | A parabola with equation <math>y=ax^2+bx+c</math> is reflected about the <math>x</math>-axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of <math>y=f(x)</math> and <math>y=g(x)</math>, respectively. Which of the following describes the graph of <math>y=(f+g)(x)?</math> |
<math> \textbf{(A)}\ \text{a parabola tangent to the }x\text{-axis} </math> | <math> \textbf{(A)}\ \text{a parabola tangent to the }x\text{-axis} </math> | ||
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: | ||
− | <cmath> \begin{align*} f(x) | + | <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> |
− | Adding them up produces: <cmath> (f + g)(x) | + | Adding them up produces: |
+ | <cmath> \begin{align*} (f + g)(x) = ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c = 20ax + 10b \end{align*}</cmath> | ||
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>. | ||
+ | |||
+ | -MP8148 | ||
==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}} |
Latest revision as of 22:55, 4 January 2019
Problem
A parabola with equation is reflected about the -axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of and , respectively. Which of the following describes the graph of
Solution 1
If we take the parabola and reflect it over the x - axis, we have the parabola . Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then:
Adding them up produces:
This is a line with slope . Since cannot be (because would be a line) we end up with
Solution 2: less computation
WLOG let the parabola be , and the reflected parabola is thus . We can also assume and Therefore, which is a non-horizontal line .
-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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