Finding the asymptotes of a rational function: linear over linear Graph all vertical and horizontal asymptotes of the rational function \[ f(x)=\frac{-2 x-11}{2 x+1} \]

Function by following steps: - step0: Find the horizontal asymptotes: \(f\left(x\right)=-2x-\frac{11}{2x+1}\) - step1: Evaluate the limits \(\lim _{x\rightarrow +\infty}\left(f\left(x\right)\right)\) and \(\lim _{x\rightarrow -\infty}\left(f\left(x\right)\right):\) \(\begin{align}&\lim _{x\rightarrow +\infty}\left(-2x-\frac{11}{2x+1}\right)\\&\lim _{x\rightarrow -\infty}\left(-2x-\frac{11}{2x+1}\right)\end{align}\) - step2: Calculate: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step3: The function has no horizontal asymptotes: \(\textrm{No horizontal asymptotes}\) Find the vertical asymptotes of \( f(x)=-2x-11/(2x+1) \). Function by following steps: - step0: Find the vertical asymptotes: \(f\left(x\right)=-2x-\frac{11}{2x+1}\) - step1: Evaluate the limit: \(\lim _{x\rightarrow -\frac{1}{2}}\left(-2x-\frac{11}{2x+1}\right)\) - step2: Evaluate the left-hand and the right-hand limits: \(\begin{align}&\lim _{x\rightarrow -\frac{1}{2}^{-}}\left(-2x-\frac{11}{2x+1}\right)\\&\lim _{x\rightarrow -\frac{1}{2}^{+}}\left(-2x-\frac{11}{2x+1}\right)\end{align}\) - step3: Evaluate the left-hand limit: \(\begin{align}&+\infty\\&-\infty\end{align}\) - step4: Limit does not exist: \(\textrm{The limit does not exist}\) - step5: \(x=-\frac{1}{2}\) is a vertical asymptote\(:\) \(\begin{align}&x=-\frac{1}{2}\textrm{ }\textrm{is a vertical asymptote}\end{align}\) - step6: List all vertical asymptotes of the function: \(\begin{align}&x=-\frac{1}{2}\end{align}\) To find the asymptotes of the rational function \[ f(x) = \frac{-2x - 11}{2x + 1} \] we have determined the following: 1. **Vertical Asymptote**: This occurs where the denominator is equal to zero. Setting the denominator \(2x + 1 = 0\) gives us: \[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \] Therefore, the vertical asymptote is at \(x = -\frac{1}{2}\). 2. **Horizontal Asymptote**: For rational functions where the degrees of the numerator and denominator are the same, the horizontal asymptote can be found by taking the ratio of the leading coefficients. In this case, the leading coefficient of the numerator is \(-2\) and that of the denominator is \(2\). Thus, we have: \[ y = \frac{-2}{2} = -1 \] However, since the degrees of the numerator and denominator are the same, we typically find a horizontal asymptote. In this case, it appears that there is no horizontal asymptote as the function approaches infinity. ### Summary of Asymptotes: - **Vertical Asymptote**: \(x = -\frac{1}{2}\) - **Horizontal Asymptote**: None ### Graphing the Asymptotes: - The vertical line \(x = -\frac{1}{2}\) will be drawn on the graph. - There is no horizontal line to graph since there is no horizontal asymptote. If you need further assistance with graphing or any other aspect, feel free to ask!