Question: If $f(x) = \frac{3}{x-2}$ and $g(x) = \sqrt{x+1}$
1. **State the problem:** We are given two functions $f(x) = \frac{3}{x-2}$ and $g(x) = \sqrt{x+1}$. We want to understand these functions and possibly analyze or graph them.
2. **Understand each function:**
- $f(x) = \frac{3}{x-2}$ is a rational function with a vertical asymptote at $x=2$ because the denominator cannot be zero.
- $g(x) = \sqrt{x+1}$ is a square root function defined only for $x \geq -1$ since the expression inside the square root must be non-negative.
3. **Domain:**
- For $f(x)$, domain is all real numbers except $x=2$.
- For $g(x)$, domain is $x \geq -1$.
4. **Key features:**
- $f(x)$ has a vertical asymptote at $x=2$ and a horizontal asymptote at $y=0$.
- $g(x)$ starts at $(-1,0)$ and increases slowly.
5. **Graphing notes:**
- $f(x)$ will approach infinity near $x=2$ from the left and negative infinity from the right.
- $g(x)$ is only defined for $x \geq -1$ and is increasing.
Final answer: The functions are $f(x) = \frac{3}{x-2}$ with domain $x \neq 2$ and $g(x) = \sqrt{x+1}$ with domain $x \geq -1$. Their graphs reflect these properties.