1. **State the problem:** We need to analyze the function $$y=\frac{(x+2)^3}{(x+1)^2}$$. 2. **Recall the formula and rules:** This is a rational function where the numerator is $(x+2)^3$ and the denominator is $(x+1)^2$. 3. **Find the domain:** The denominator cannot be zero, so $x+1 \neq 0 \Rightarrow x \neq -1$. 4. **Find intercepts:** - **x-intercept(s):** Set numerator equal to zero: $(x+2)^3=0 \Rightarrow x=-2$. - **y-intercept:** Set $x=0$: $$y=\frac{(0+2)^3}{(0+1)^2}=\frac{8}{1}=8$$. 5. **Simplify and analyze behavior:** No common factors to cancel between numerator and denominator. 6. **Vertical asymptote:** At $x=-1$ because denominator is zero and numerator is not zero there. 7. **Horizontal or oblique asymptote:** Degree numerator = 3, degree denominator = 2, so no horizontal asymptote; instead, there is an oblique asymptote. 8. **Find oblique asymptote by polynomial division:** Divide $(x+2)^3$ by $(x+1)^2$. Expand numerator: $$(x+2)^3 = x^3 + 6x^2 + 12x + 8$$ Expand denominator: $$(x+1)^2 = x^2 + 2x + 1$$ Divide: $$\frac{x^3 + 6x^2 + 12x + 8}{x^2 + 2x + 1}$$ First term: $x^3 \div x^2 = x$ Multiply back: $x(x^2 + 2x + 1) = x^3 + 2x^2 + x$ Subtract: $$(x^3 + 6x^2 + 12x + 8) - (x^3 + 2x^2 + x) = 4x^2 + 11x + 8$$ Next term: $4x^2 \div x^2 = 4$ Multiply back: $4(x^2 + 2x + 1) = 4x^2 + 8x + 4$ Subtract: $$(4x^2 + 11x + 8) - (4x^2 + 8x + 4) = 3x + 4$$ So quotient is $x + 4$ with remainder $3x + 4$. Oblique asymptote: $$y = x + 4$$ 9. **Summary:** - Domain: $x \neq -1$ - x-intercept: $x = -2$ - y-intercept: $y = 8$ - Vertical asymptote: $x = -1$ - Oblique asymptote: $y = x + 4$ This completes the analysis of the function.