a. Find any intercepts. b. Find any vertical, horizontal, and/or slant (oblique) asymptotes \( f(x)=\frac{x^{3}+8}{x^{2}+6 x} \) s. Inere are no \( x \)-intercepis. b. Find the vertical asymptote(s). Select the correct choice below and, if necessary, fill in the answer box to complete the choice. A. The equation of the vertical asymptote(s) are B. There is no vertical asymptote. Find the horizontal asymptote(s). Select the correct choice below and, if necessary, fill in the answer box to complete the choice. A. The equation of the horizontal asymptote is (Type an equation.) B. There is no horizontal asymptote. Find the slant (oblique) asymptote(s). Select the correct choice below and, if necessary, fill in the answer box to complete the choice. A. The equation of the slant asymptote is (Type an equation.) B. There is no slant asymptote.

Let's analyze the function \( f(x) = \frac{x^{3} + 8}{x^{2} + 6x} \) step by step to find its intercepts and asymptotes. ### a. Intercepts **1. \( x \)-Intercept(s):** To find the \( x \)-intercept(s), set \( f(x) = 0 \): \[ \frac{x^{3} + 8}{x^{2} + 6x} = 0 \implies x^{3} + 8 = 0 \implies x^{3} = -8 \implies x = -2 \] **\( x \)-Intercept:** \( (-2, 0) \) **2. \( y \)-Intercept:** To find the \( y \)-intercept, set \( x = 0 \): \[ f(0) = \frac{0^{3} + 8}{0^{2} + 6(0)} = \frac{8}{0} \] The function is undefined at \( x = 0 \), so there is **no \( y \)-intercept**. ### b. Vertical Asymptote(s) Vertical asymptotes occur where the denominator is zero (and the numerator is not zero). Set the denominator equal to zero: \[ x^{2} + 6x = 0 \implies x(x + 6) = 0 \implies x = 0 \text{ or } x = -6 \] Check the numerator at these points: - At \( x = 0 \): \( x^{3} + 8 = 8 \neq 0 \) - At \( x = -6 \): \( (-6)^{3} + 8 = -208 \neq 0 \) **Conclusion:** There are vertical asymptotes at: \[ x = 0 \quad \text{and} \quad x = -6 \] **Choice A:** The equation of the vertical asymptote(s) are \( x = 0 \) and \( x = -6 \). ### c. Horizontal Asymptote(s) Compare the degrees of the numerator and the denominator: - **Numerator Degree:** 3 - **Denominator Degree:** 2 Since the degree of the numerator is **greater** than the degree of the denominator by 1, there is **no horizontal asymptote**. Instead, there is a **slant (oblique) asymptote**. **Choice B:** There is no horizontal asymptote. ### d. Slant (Oblique) Asymptote To find the slant asymptote, perform polynomial long division of the numerator by the denominator. Divide \( x^{3} + 8 \) by \( x^{2} + 6x \): 1. **First Division Step:** \[ \frac{x^{3}}{x^{2}} = x \] Multiply the divisor by \( x \): \[ x(x^{2} + 6x) = x^{3} + 6x^{2} \] Subtract from the original polynomial: \[ (x^{3} + 0x^{2} + 0x + 8) - (x^{3} + 6x^{2}) = -6x^{2} + 0x + 8 \] 2. **Second Division Step:** \[ \frac{-6x^{2}}{x^{2}} = -6 \] Multiply the divisor by \( -6 \): \[ -6(x^{2} + 6x) = -6x^{2} - 36x \] Subtract: \[ (-6x^{2} + 0x + 8) - (-6x^{2} - 36x) = 36x + 8 \] The quotient is \( x - 6 \) with a remainder of \( 36x + 8 \). Therefore, the **slant asymptote** is: \[ y = x - 6 \] **Choice A:** The equation of the slant asymptote is \( y = x - 6 \). ### Summary of Answers - **a. Intercepts:** - \( x \)-intercept at \( (-2, 0) \) - No \( y \)-intercept - **b. Vertical Asymptotes:** - \( x = 0 \) and \( x = -6 \) - **c. Horizontal Asymptote:** - **There is no horizontal asymptote.** - **d. Slant (Oblique) Asymptote:** - \( y = x - 6 \)