a) \( f(x)=\frac{x+1}{x-2} \)

The function you've provided is \[ f(x) = \frac{x+1}{x-2}. \] This is a rational function, which means it is the ratio of two polynomials. Here are some key features of this function: 1. **Domain**: The function is defined for all \( x \) except where the denominator is zero. To find the values that make the denominator zero, set \( x - 2 = 0 \). Thus, \( x = 2 \) is not in the domain. Therefore, the domain of \( f(x) \) is: \[ \text{Domain: } x \in \mathbb{R}, x \neq 2. \] 2. **Vertical Asymptote**: Since the function is undefined at \( x = 2 \), there is a vertical asymptote at this point. 3. **Horizontal Asymptote**: To find the horizontal asymptote, we look at the behavior of \( f(x) \) as \( x \) approaches infinity. The degrees of the numerator and denominator are the same (both are degree 1). The horizontal asymptote can be found by taking the ratio of the leading coefficients: \[ y = \frac{1}{1} = 1. \] So, there is a horizontal asymptote at \( y = 1 \). 4. **Intercepts**: - **Y-intercept**: To find the y-intercept, set \( x = 0 \): \[ f(0) = \frac{0 + 1}{0 - 2} = \frac{1}{-2} = -\frac{1}{2}. \] So, the y-intercept is \( (0, -\frac{1}{2}) \). - **X-intercept**: To find the x-intercept, set \( f(x) = 0 \): \[ \frac{x + 1}{x - 2} = 0 \implies x + 1 = 0 \implies x = -1. \] So, the x-intercept is \( (-1, 0) \). 5. **Behavior near the asymptotes**: - As \( x \) approaches 2 from the left (\( x \to 2^- \)), \( f(x) \to -\infty \). - As \( x \) approaches 2 from the right (\( x \to 2^+ \)), \( f(x) \to +\infty \). - As \( x \to \infty \), \( f(x) \to 1 \). - As \( x \to -\infty \), \( f(x) \to 1 \). 6. **Graph**: The graph of this function will show a hyperbolic shape with the vertical asymptote at \( x = 2 \) and the horizontal asymptote at \( y = 1 \). If you have any specific questions or need further analysis on this function, feel free to ask!