1. **State the problem:** Simplify or understand the function $$y=\frac{x+1}{x-1}$$. 2. This is a rational function where numerator is $$x+1$$ and denominator is $$x-1$$. 3. The vertical asymptote is at $$x=1$$ (denominator zero). 4. The horizontal asymptote is $$y=1$$ as $$x \to \pm \infty$$ because coefficients of highest degree terms in numerator and denominator are both 1. --- 1. **State the problem:** Simplify $$y = (1-x)(+x)^{-1}$$. 2. Interpreting the expression carefully, assume it means $$y = (1-x)x^{-1} = \frac{1-x}{x}$$. 3. Rewrite as $$y = \frac{1}{x} - 1$$. --- 1. **State the problem:** Simplify $$u = \frac{r}{x} = \frac{0}{r}x^{-\frac{5}{3}}$$. 2. The left side is $$u = \frac{r}{x}$$. 3. The right side is $$\frac{0}{r}x^{-\frac{5}{3}} = 0$$ since numerator zero. 4. Potentially inconsistent, but the term $$\frac{0}{r}x^{-\frac{5}{3}}$$ always zero. --- 1. **State the problem:** Simplify $$y = \sqrt{x} \left(5 - x - \frac{x^2}{r}\right)$$. 2. Express square root as $$x^{\frac{1}{2}}$$. 3. Distribute: $$y = x^{\frac{1}{2}}\cdot 5 - x^{\frac{3}{2}} - x^{\frac{5}{2}}/r$$. --- 1. **State the problem:** Simplify $$s(t) = \frac{1 + \sqrt{t}}{1 - \sqrt{t}}$$. 2. Let $$u = \sqrt{t}$$ to get $$s(u) = \frac{1+u}{1-u}$$. 3. Recognize this is a rational function with vertical asymptote at $$u=1$$ or $$t=1$$. --- 1. **State the problem:** Analyze $$g(y) = \frac{2}{1-y^4}$$. 2. The denominator is zero when $$1 - y^4 = 0 \Rightarrow y^4 = 1 \Rightarrow y = \pm 1$$. 3. Vertical asymptotes at $$y=1$$ and $$y=-1$$. ---