For each function, find the domain and the vertical and horizontal asymptotes (if any). \( g(x)=\frac{x+2}{x-5} \)

Function by following steps: - step0: Find the horizontal asymptotes: \(g\left(x\right)=\frac{x+2}{x-5}\) - step1: Evaluate the limits \(\lim _{x\rightarrow +\infty}\left(g\left(x\right)\right)\) and \(\lim _{x\rightarrow -\infty}\left(g\left(x\right)\right):\) \(\begin{align}&\lim _{x\rightarrow +\infty}\left(\frac{x+2}{x-5}\right)\\&\lim _{x\rightarrow -\infty}\left(\frac{x+2}{x-5}\right)\end{align}\) - step2: Calculate: \(\begin{align}&1\\&1\end{align}\) - step3: The finite values are horizontal asymptotes: \(\begin{align}&g\left(x\right)=1\end{align}\) Find the vertical asymptotes of \( g(x)=\frac{x+2}{x-5} \). Function by following steps: - step0: Find the vertical asymptotes: \(g\left(x\right)=\frac{x+2}{x-5}\) - step1: Evaluate the limit: \(\lim _{x\rightarrow 5}\left(\frac{x+2}{x-5}\right)\) - step2: Evaluate the left-hand and the right-hand limits: \(\begin{align}&\lim _{x\rightarrow 5^{-}}\left(\frac{x+2}{x-5}\right)\\&\lim _{x\rightarrow 5^{+}}\left(\frac{x+2}{x-5}\right)\end{align}\) - step3: Evaluate the left-hand limit: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step4: Limit does not exist: \(\textrm{The limit does not exist}\) - step5: \(x=5\) is a vertical asymptote\(:\) \(\begin{align}&x=5\textrm{ }\textrm{is a vertical asymptote}\end{align}\) - step6: List all vertical asymptotes of the function: \(\begin{align}&x=5\end{align}\) Let's analyze the function \( g(x) = \frac{x+2}{x-5} \) step by step to find its domain and asymptotes. ### Step 1: Finding the Domain The domain of a function is the set of all possible input values (x-values) for which the function is defined. For rational functions, the function is undefined where the denominator is zero. 1. Set the denominator equal to zero: \[ x - 5 = 0 \] Solving this gives: \[ x = 5 \] Thus, the function is undefined at \( x = 5 \). Therefore, the domain of \( g(x) \) is all real numbers except \( x = 5 \): \[ \text{Domain: } (-\infty, 5) \cup (5, \infty) \] ### Step 2: Finding Vertical Asymptotes Vertical asymptotes occur where the function approaches infinity, which happens at the values that make the denominator zero (and are not canceled by the numerator). From our previous calculation, we found that: \[ x = 5 \] is where the function is undefined. Therefore, there is a vertical asymptote at: \[ x = 5 \] ### Step 3: Finding Horizontal Asymptotes Horizontal asymptotes describe the behavior of the function as \( x \) approaches infinity or negative infinity. For rational functions, the horizontal asymptote can be determined by comparing the degrees of the polynomial in the numerator and the denominator. In this case: - The degree of the numerator \( (x + 2) \) is 1. - The degree of the denominator \( (x - 5) \) is also 1. When the degrees are equal, the horizontal asymptote is given by the ratio of the leading coefficients. Here, both leading coefficients are 1. Thus, the horizontal asymptote is: \[ y = 1 \] ### Summary - **Domain:** \( (-\infty, 5) \cup (5, \infty) \) - **Vertical Asymptote:** \( x = 5 \) - **Horizontal Asymptote:** \( y = 1 \)