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>\mathrm{E}</math>.
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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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 19:48, 13 June 2015

Problem

The parabola with equation $p(x) = ax^2+bx+c$ and vertex $(h,k)$ is reflected about the line $y=k$. This results in the parabola with equation $q(x) = dx^2+ex+f$. Which of the following equals $a+b+c+d+e+f$?

$(\mathrm{A})\ 2b \qquad (\mathrm{B})\ 2c \qquad (\mathrm{C})\ 2a+2b \qquad (\mathrm{D})\ 2h \qquad (\mathrm{E})\ 2k$

Solution

We write $p(x)$ as $a(x-h)^2+k$ (this is possible for any parabola). Then the reflection of $p(x)$ is $q(x) = -a(x-h)^2+k$. Then we find $p(x) + q(x) = 2k$. Since $p(1) = a+b+c$ and $q(1) = d+e+f$, we have $a+b+c+d+e+f = 2k$, so the answer is $\fbox{E}$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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

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