The graph of which of the following functions has exactly one horizontal asymptote and no vertical asymptotes? (A) \( y=\frac{1}{x^{2}+1} \) (B) \( y=\frac{1}{x^{3}+1} \) (C) \( y=\frac{1}{e^{z}-1} \) (D) \( y=\frac{1}{e^{x}+1} \)

To solve this, let's analyze the behavior of each function as \( x \) approaches infinity and near points where the denominator could become zero (indicating vertical asymptotes). (A) \( y=\frac{1}{x^{2}+1} \) approaches \( 0 \) as \( x \) approaches \( \pm \infty \), but there are no vertical asymptotes since \( x^2 + 1 > 0 \) for all \( x \). (B) \( y=\frac{1}{x^{3}+1} \) also approaches \( 0 \) as \( x \) approaches \( \pm \infty \), but again, there are no vertical asymptotes because \( x^3 + 1 \neq 0 \) for all \( x \). (C) \( y=\frac{1}{e^{z}-1} \) has a vertical asymptote at \( z = 0 \) because \( e^0 - 1 = 0 \). (D) \( y=\frac{1}{e^{x}+1} \) approaches \( 0 \) as \( x \) approaches \( \infty \) and \( 1 \) as \( x \) approaches \( -\infty \). There are no vertical asymptotes since \( e^x + 1 > 0 \) for all \( x \). Thus, the function with exactly one horizontal asymptote and no vertical asymptotes is \( (A) \) \( y = \frac{1}{x^2 + 1} \).