1. **State the problem:** We are given the function $$f(x) = \frac{x^3 + 5x^2 - 6x}{x^2 - 1}$$ and need to find: (i) The vertical asymptote(s) and/or hole(s) of the graph. (ii) The horizontal or oblique asymptote(s), if any. 2. **Recall important rules:** - Vertical asymptotes occur where the denominator is zero but the numerator is not zero at those points. - Holes occur where both numerator and denominator are zero at the same point (common factors). - Horizontal asymptotes depend on the degrees of numerator and denominator polynomials. - Oblique (slant) asymptotes occur if the degree of numerator is exactly one more than denominator. 3. **Find vertical asymptotes and holes:** - Factor numerator and denominator: $$x^3 + 5x^2 - 6x = x(x^2 + 5x - 6) = x(x+6)(x-1)$$ $$x^2 - 1 = (x-1)(x+1)$$ - Simplify the function: $$f(x) = \frac{x(x+6)(x-1)}{(x-1)(x+1)} = \frac{x(x+6)\cancel{(x-1)}}{\cancel{(x-1)}(x+1)} = \frac{x(x+6)}{x+1}, \quad x \neq 1$$ - The factor $(x-1)$ cancels, so at $x=1$ there is a hole. - The denominator is zero at $x = -1$ and numerator is not zero there, so vertical asymptote at $x = -1$. 4. **Find horizontal or oblique asymptotes:** - Degree numerator after simplification: 2 (since numerator is $x(x+6) = x^2 + 6x$) - Degree denominator: 1 - Since numerator degree (2) is exactly one more than denominator degree (1), there is an oblique asymptote. - Perform polynomial division of numerator by denominator: $$\frac{x^2 + 6x}{x+1}$$ Divide: - $x^2 \div x = x$, multiply $x(x+1) = x^2 + x$ - Subtract: $(x^2 + 6x) - (x^2 + x) = 5x$ - Divide: $5x \div x = 5$, multiply $5(x+1) = 5x + 5$ - Subtract: $5x - (5x + 5) = -5$ - Remainder $-5$ over divisor $x+1$ - So quotient is $x + 5$ and remainder $-5/(x+1)$ - Oblique asymptote is the line: $$y = x + 5$$ **Final answers:** - Vertical asymptote at $x = -1$ - Hole at $x = 1$ - Oblique asymptote: $y = x + 5$