Questions: Identify all asymptotes of f(x)=(x^2+2x-15)/(x^3+1) A. vertical asymptote: x=-1; horizontal asymptotes: y=-5 and γ=3 B. no vertical asymptote; horizontal asymptote: γ=0 C. no vertical asymptote; horizontal asymptote: γ=-1 D. vertical asymptote: x=-1; horizontal asymptote: γ=0

Identify all asymptotes of f(x)=(x^2+2x-15)/(x^3+1) A. vertical asymptote: x=-1; horizontal asymptotes: y=-5 and γ=3 B. no vertical asymptote; horizontal asymptote: γ=0 C. no vertical asymptote; horizontal asymptote: γ=-1 D. vertical asymptote: x=-1; horizontal asymptote: γ=0
Transcript text: Identify all asymptotes of $f(x)=\frac{x^{2}+2 x-15}{x^{3}+1}$ A. vertical asymptote: $x=-1$; horizontal asymptotes: $y=-5$ and $\gamma=3$ B. no vertical asymptote; horizontal asymptote: $\gamma=0$ C. no vertical asymptote; horizontal asymptote: $\gamma=-1$ D. vertical asymptote: $x=-1$; horizontal asymptote: $\gamma=0$
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Solution

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Solution Steps

Step 1: Finding Vertical Asymptotes

To find the vertical asymptotes, we set the denominator \( x^3 + 1 = 0 \). Solving this gives us the roots: \[ x = -1, \quad x = \frac{1}{2} - \frac{\sqrt{3}}{2}i, \quad x = \frac{1}{2} + \frac{\sqrt{3}}{2}i \] Since vertical asymptotes occur where the denominator is zero and the numerator is non-zero, we only consider the real solution \( x = -1 \) as the vertical asymptote.

Step 2: Finding Horizontal Asymptotes

Next, we determine the horizontal asymptote by comparing the degrees of the numerator and denominator. The degree of the numerator \( x^2 + 2x - 15 \) is 2, and the degree of the denominator \( x^3 + 1 \) is 3. Since the degree of the numerator is less than that of the denominator, the horizontal asymptote is given by: \[ y = 0 \]

Final Answer

The function \( f(x) = \frac{x^2 + 2x - 15}{x^3 + 1} \) has a vertical asymptote at \( x = -1 \) and a horizontal asymptote at \( y = 0 \). Therefore, the answer is: \[ \boxed{\text{D: vertical asymptote: } x = -1; \text{ horizontal asymptote: } y = 0} \]

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