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) \( y=\frac{1}{e^{x}+1} \)

We analyze each function one by one. 1. For \( y=\frac{1}{x^{2}+1} \): - **Domain:** The denominator \( x^2+1 \) is always positive since \( x^2 \geq 0 \), so \( x^2+1 \geq 1 \). Thus, there are no \( x \) for which the denominator is zero. There are no vertical asymptotes. - **Horizontal asymptote:** As \( x \to \pm\infty \), \( x^2+1 \to \infty \) so \[ y \to \frac{1}{\infty} = 0. \] There is exactly one horizontal asymptote at \( y=0 \). 2. For \( y=\frac{1}{x^{3}+1} \): - **Domain:** The denominator \( x^3+1 \) equals zero when \[ x^3+1 = 0 \quad \Rightarrow \quad x = -1. \] Therefore, there is a vertical asymptote at \( x=-1 \). 3. For \( y=\frac{1}{e^{z}-1} \): - **Domain:** Here the variable is \( z \). The denominator \( e^z-1 \) equals zero when \[ e^z-1=0 \quad \Rightarrow \quad z=0. \] This produces a vertical asymptote at \( z=0 \). 4. For \( y=\frac{1}{e^{x}+1} \): - **Domain:** Since \( e^x>0 \) for all \( x \), \( e^x+1>1 \) and no division by zero occurs. - **Horizontal asymptotes:** - As \( x\to\infty \), \( y\to \frac{1}{\infty+1} = 0 \). - As \( x\to -\infty \), \( e^x\to 0 \), so \( y\to\frac{1}{0+1} = 1 \). This gives two horizontal asymptotes, \( y=0 \) and \( y=1 \). The function that meets the criteria of exactly one horizontal asymptote and no vertical asymptotes is \[ \boxed{y=\frac{1}{x^{2}+1}}. \]