Difference between revisions of "2003 AMC 12A Problems/Problem 19"
Epicfailiure (talk | contribs) (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…') |
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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: | + | 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) | + | <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: | + | 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> |
− | + | 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 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. Unknown error_msg)
This is a line with slope . Since
cannot be
(because
would be a line) we end up with