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  1. I'll assume n>1. For large values of x, k^n becomes negligible compared to x^n and we get: y = Ax^n/x^n = A So y=A is the horizontal asymptote. When y = A/2: A/2= Ax^n/(k^n+x^n) k^n+x^n = 2x^n k^n = x^n x = k

    1 Answers · Science & Mathematics · 13/09/2020

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