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- The graph of which of the following functions has exactly one horizontal asymptote and no vertical asymptotes?
A) y = (1)/(x^2 + 1)
B) y = (1)/(x^3 + 1)
C) y = (1)/(e^x + 1)
D) y = (1)/(x^2 – 1)
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The Correct Answer and Explanation is:
The correct answer is C) y = (1)/(e^x + 1).
Explanation:
To determine which function has exactly one horizontal asymptote and no vertical asymptotes, we analyze each option:
1. Horizontal Asymptotes:
A function has a horizontal asymptote when the limit of f(x)f(x) as xx approaches infinity or negative infinity tends toward a specific finite value.
- Option A: y=1×2+1y = \frac{1}{x^2 + 1}
- As x→±∞x \to \pm\infty, x2x^2 grows indefinitely, meaning x2+1→∞x^2 + 1 \to \infty, and thus y→0y \to 0.
- There is one horizontal asymptote at y=0y = 0.
- However, this function has no vertical asymptotes because the denominator never equals zero.
- Option B: y=1×3+1y = \frac{1}{x^3 + 1}
- For large xx, x3+1x^3 + 1 dominates, leading to y→0y \to 0.
- One horizontal asymptote exists at y=0y = 0, and there are no vertical asymptotes since the denominator never vanishes.
- Option C: y=1ex+1y = \frac{1}{e^x + 1}
- As x→∞x \to \infty, exe^x grows exponentially, meaning ex+1→∞e^x + 1 \to \infty and thus y→0y \to 0.
- As x→−∞x \to -\infty, exe^x approaches 0, making y=11=1y = \frac{1}{1} = 1.
- Two horizontal asymptotes exist at y=0y = 0 and y=1y = 1, eliminating this option.
- Option D: y=1×2−1y = \frac{1}{x^2 – 1}
- The denominator equals zero when x2=1x^2 = 1, meaning vertical asymptotes exist at x=±1x = \pm1.
- This contradicts the given condition.
Thus, Option C is the correct answer because it has exactly one horizontal asymptote and no vertical asymptotes.
