2024 AMC 12A Problem 13The graph of y=ex+1+e−x−2y=e^{x+1}+e^{-x}-2y=ex+1+e−x−2 has an axis of symmetry. What is the reflection of the point (−1,12)\left(-1,\frac{1}{2}\right)(−1,21) over this axis?(A) (−1,−32)\text{(A)}\;\left(-1,-\frac{3}{2}\right)(A)(−1,−23)(B) (−1,0)\text{(B)}\;(-1,0)(B)(−1,0)(C) (−1,12)\text{(C)}\;\left(-1,\frac{1}{2}\right)(C)(−1,21)(D) (0,12)\text{(D)}\;\left(0,\frac{1}{2}\right)(D)(0,21)(E) (3,12)\text{(E)}\;\left(3,\frac{1}{2}\right)(E)(3,21)Related TopicsCoreGeometric TransformationsCheck AnswerYour answer:ABCDECheckHints (3)Hint 1Assumef(x)=ex+1+e−x−2f(x)=e^{x+1}+e^{-x}-2f(x)=ex+1+e−x−2and has axis of symmetryx=ax=ax=aHint 2f(2a−x)=f(x)f(2a-x)=f(x)f(2a−x)=f(x)e2a−x+1+e−2a+x−2=ex+1+e−x−2e^{2a-x+1}+e^{-2a+x}-2=e^{x+1}+e^{-x}-2e2a−x+1+e−2a+x−2=ex+1+e−x−2e2a−x+1‾+e−2a+x‾‾=ex+1‾‾+e−x‾\underline{e^{2a-x+1}}+\underline{\underline{e^{-2a+x}}}=\underline{\underline{e^{x+1}}}+\underline{e^{-x}}e2a−x+1+e−2a+x=ex+1+e−x2a−x+1=−x2a-x+1=-x2a−x+1=−x⇒2a+1=0\Rightarrow 2a+1=0⇒2a+1=0⇒a=−12\Rightarrow a=-\frac{1}{2}⇒a=−21−2a+x=x+1-2a+x=x+1−2a+x=x+1⇒2a+1=0\Rightarrow 2a+1=0⇒2a+1=0⇒a=−12\Rightarrow a=-\frac{1}{2}⇒a=−21Hint 3x0=2(−12)−(−1)=−1+1=0x_0=2\left(-\frac{1}{2}\right)-(-1)=-1+1=0x0=2(−21)−(−1)=−1+1=0P′(0,12)P'\left(0,\frac{1}{2}\right)P′(0,21)Related Problems (2)2024 AMC 10A Problem 132024 AMC 12A Problem 25Final Answer(D) (0,12)\left(0,\frac{1}{2}\right)(0,21)
