Difference between revisions of "2024 AMC 12A Problems/Problem 13"
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− | image by ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | + | image by ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] inspired from Kevin_liu |
==Solution 6 == | ==Solution 6 == | ||
− | <cmath> f(-x-1) = e^{-x-1+1} + e^{-(-x-1)} - 2 = e^{-x} + e^{x+1 | + | <cmath> f(-x-1) = e^{-x-1+1} + e^{-(-x-1)} - 2 = e^{-x} + e^{x+1} - 2 = f(x) </cmath> |
so the axis is middle point of (x , -x-1 ) = -1/2 | so the axis is middle point of (x , -x-1 ) = -1/2 | ||
point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D | point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D | ||
+ | |||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
Revision as of 17:02, 9 November 2024
Contents
[hide]Problem
The graph of has an axis of symmetry. What is the reflection of the point over this axis?
Solution 1
The line of symmetry is probably of the form for some constant . A vertical line of symmetry at for a function exists if and only if ; we substitute and into our given function and see that we must have
for all real . Simplifying:
If , then for all real ; this is clearly impossible, so let . Thus, our line of symmetry is , and reflecting over this line gives
~Technodoggo
Solution 2 (Graphing cheese)
Consider the graphs of and . A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus, is the only possible answer.
Note: You can more rigorously think about the solution by noting that since the derivative of the power that e is raised to in one equation is equal to the derivative of the power that e is raised to multiplied by and both equations are subtracted by 1, then the sum of both equations will be the same from one side of the interception to the other. Setting both equations equal to each other, it is trivial to see , giving us the axis of symmetry.
~woeIsMe
(graph by Technodoggo)
Solution 3(Derivatives)
For an axis of symmetry there must be a local maximum or minimum. I mean, think about it. Would it make sense for a function to be larger than to the left of its axis of symmetry? So we do a little derivativing and set said derivative equal to 0. Now in order to solve for e we do a lil subbing hehehe -> -> -> -> -> -> -> Now we just flippidy flap (-1, ) over and blammo, we got ourselves the answer choice
Note: "an axis of symmetry there must be a local maximum or minimum" is not true. consider the graph of 1/x
Solution 4
so f(0) = f(-1) , f(1) =f(-2)
then f(x) is symmetric about x=-1/2
point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D
~luckuso
Solution 5
Notice that this is simply a transformation of , specifically . We know that has a parabolic shape with an axis of symmetry . Since this is simple a transformation of to the left of the origin, we have the axis of symmetry is giving us
~KEVIN_LIU
Solution 5a (cosh(x) image )
image by ~luckuso inspired from Kevin_liu
Solution 6
so the axis is middle point of (x , -x-1 ) = -1/2
point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D
Solution 7
symmetric on axis a means f(2a-x) = f(x)
plug in 2a+1 = 0 solve a= -1/2
point (-1, 1/2) reflects over axis x=-1/2 is point ( 0, 1/2) answer choice D ~luckuso
See also
2024 AMC 12A (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.