Difference between revisions of "2003 AMC 12A Problems/Problem 19"
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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: | 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) &= ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c &= 20ax + 10b </cmath> | 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> |
Revision as of 20:31, 20 June 2015
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
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:
\[(f + g)(x) &= ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c &= 20ax + 10b\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
This is a line with slope . Since cannot be (because would be a line) we end up with
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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