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2024 AMC 12A Problem 13

Problem 13 of 25IntermediateAlgebra

The graph of y=ex+1+ex2y=e^{x+1}+e^{-x}-2 has an axis of symmetry. What is the reflection of the point (1,12)\left(-1,\tfrac12\right) over this axis?

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Solution

The curve y=ex+1+ex2y=e^{x+1}+e^{-x}-2 is symmetric about the vertical line through its minimum. Setting the derivative ex+1ex=0e^{x+1}-e^{-x}=0 gives x+1=x,x+1=-x, so x=12.x=-\tfrac12. Reflecting (1,12)\left(-1,\tfrac12\right) across x=12x=-\tfrac12 keeps the yy-coordinate and sends x=1x=-1 to x=0.x=0. The image is (0,12).\left(0,\tfrac12\right). Thus, the correct answer is D.

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Concepts: function · symmetry

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.