Difference between revisions of "2001 AMC 12 Problems/Problem 13"
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== Solution == | == Solution == | ||
We write <math>p(x)</math> as <math>a(x-h)^2+k</math> (this is possible for any parabola). Then the reflection of <math>p(x)</math> is <math>q(x) = -a(x-h)^2+k</math>. Then we find <math>p(x) + q(x) = 2k</math>. Since <math>p(1) = a+b+c</math> and <math>q(1) = d+e+f</math>, we have <math>a+b+c+d+e+f = 2k</math>, so the answer is <math>\fbox{E}</math>. | We write <math>p(x)</math> as <math>a(x-h)^2+k</math> (this is possible for any parabola). Then the reflection of <math>p(x)</math> is <math>q(x) = -a(x-h)^2+k</math>. Then we find <math>p(x) + q(x) = 2k</math>. Since <math>p(1) = a+b+c</math> and <math>q(1) = d+e+f</math>, we have <math>a+b+c+d+e+f = 2k</math>, so the answer is <math>\fbox{E}</math>. | ||
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| + | == Video Solution == | ||
| + | https://youtu.be/ZbHZTiljP1o | ||
== See Also == | == See Also == | ||
Latest revision as of 21:22, 24 October 2021
Contents
Problem
The parabola with equation 
and vertex 
is reflected about the line 
. This results in the parabola with equation 
. Which of the following equals 
?
Solution
We write 
as 
(this is possible for any parabola). Then the reflection of is 
. Then we find 
. Since 
and 
, we have 
, so the answer is 
.
Video Solution
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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. 
