1. We are asked to find the equations of the horizontal and vertical asymptotes of the given irrational functions. 2. Recall that vertical asymptotes occur where the denominator of a function approaches zero and the function tends to infinity, while horizontal asymptotes describe the behavior of the function as $x \to \pm \infty$. 3. For function a: $$f(x) = \sqrt{\frac{x+2}{x^2 - 3x - 4}}$$ - Factor the denominator: $$x^2 - 3x - 4 = (x - 4)(x + 1)$$ - Vertical asymptotes occur where denominator is zero and numerator is not zero: $$x - 4 = 0 \Rightarrow x = 4$$ $$x + 1 = 0 \Rightarrow x = -1$$ - Check domain restrictions: numerator $x+2$ is zero at $x = -2$, which is not a vertical asymptote. - Horizontal asymptote: analyze limit as $x \to \pm \infty$: $$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \sqrt{\frac{x+2}{x^2 - 3x - 4}}$$ Since degree of denominator (2) > degree of numerator (1), the fraction tends to zero. So, $$\lim_{x \to \pm \infty} f(x) = 0$$ Hence, horizontal asymptote is: $$y = 0$$ 4. For function b: $$f(x) = \sqrt{\frac{4x^3 + 32}{x^3 + 3x^2 + 3x + 1}}$$ - Factor denominator if possible: $$x^3 + 3x^2 + 3x + 1 = (x+1)^3$$ - Vertical asymptote occurs where denominator is zero and numerator is not zero: $$x + 1 = 0 \Rightarrow x = -1$$ - Check numerator at $x = -1$: $$4(-1)^3 + 32 = -4 + 32 = 28 \neq 0$$ So vertical asymptote at $x = -1$. - Horizontal asymptote: analyze limit as $x \to \pm \infty$: $$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \sqrt{\frac{4x^3 + 32}{x^3 + 3x^2 + 3x + 1}} = \lim_{x \to \pm \infty} \sqrt{\frac{4 + \frac{32}{x^3}}{1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}}} = \sqrt{4} = 2$$ So horizontal asymptote is: $$y = 2$$ 5. For function c: $$f(x) = \sqrt[3]{\frac{3x^2 - x}{6x^2 - 5x + 1}}$$ - Vertical asymptotes occur where denominator is zero and numerator is not zero. - Factor denominator: $$6x^2 - 5x + 1$$ Use quadratic formula: $$x = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12}$$ So roots: $$x = \frac{6}{12} = \frac{1}{2}, \quad x = \frac{4}{12} = \frac{1}{3}$$ - Check numerator at these points: $$3x^2 - x = x(3x - 1)$$ At $x=\frac{1}{2}$: $$3(\frac{1}{2})^2 - \frac{1}{2} = 3 \cdot \frac{1}{4} - \frac{1}{2} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \neq 0$$ At $x=\frac{1}{3}$: $$3(\frac{1}{3})^2 - \frac{1}{3} = 3 \cdot \frac{1}{9} - \frac{1}{3} = \frac{1}{3} - \frac{1}{3} = 0$$ So at $x=\frac{1}{3}$ numerator is zero, so no vertical asymptote there. Vertical asymptote only at $x=\frac{1}{2}$. - Horizontal asymptote: analyze limit as $x \to \pm \infty$: $$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \sqrt[3]{\frac{3x^2 - x}{6x^2 - 5x + 1}} = \sqrt[3]{\frac{3 - \frac{1}{x}}{6 - \frac{5}{x} + \frac{1}{x^2}}} = \sqrt[3]{\frac{3}{6}} = \sqrt[3]{\frac{1}{2}}$$ So horizontal asymptote is: $$y = \sqrt[3]{\frac{1}{2}}$$ Summary: - Function a vertical asymptotes: $x = 4$, $x = -1$; horizontal asymptote: $y = 0$ - Function b vertical asymptote: $x = -1$; horizontal asymptote: $y = 2$ - Function c vertical asymptote: $x = \frac{1}{2}$; horizontal asymptote: $y = \sqrt[3]{\frac{1}{2}}$