Find the slant asymptote and the vertical asymptote. (If an answer does not exist, enter DNE.) \[ r(x)=\frac{9 x-x^{2}}{8 x-8} \] slant asymptote vertical asymptote

Function by following steps: - step0: Find the vertical asymptotes: \(r\left(x\right)=\frac{9x-x^{2}}{8x-8}\) - step1: Evaluate the limit: \(\lim _{x\rightarrow 1}\left(\frac{9x-x^{2}}{8x-8}\right)\) - step2: Evaluate the left-hand and the right-hand limits: \(\begin{align}&\lim _{x\rightarrow 1^{-}}\left(\frac{9x-x^{2}}{8x-8}\right)\\&\lim _{x\rightarrow 1^{+}}\left(\frac{9x-x^{2}}{8x-8}\right)\end{align}\) - step3: Evaluate the left-hand limit: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step4: Limit does not exist: \(\textrm{The limit does not exist}\) - step5: \(x=1\) is a vertical asymptote\(:\) \(\begin{align}&x=1\textrm{ }\textrm{is a vertical asymptote}\end{align}\) - step6: List all vertical asymptotes of the function: \(\begin{align}&x=1\end{align}\) Find the oblique asymptotes of \( r(x)=\frac{9 x-x^{2}}{8 x-8} \). Function by following steps: - step0: Find the oblique asymptotes: \(r\left(x\right)=\frac{9x-x^{2}}{8x-8}\) - step1: Evaluate the limits \(\lim _{x\rightarrow +\infty}\left(\frac{r\left(x\right)}{x}\right)\) and \(\lim _{x\rightarrow -\infty}\left(\frac{r\left(x\right)}{x}\right):\) \(\begin{align}&\lim _{x\rightarrow +\infty}\left(\frac{\frac{9x-x^{2}}{8x-8}}{x}\right)\\&\lim _{x\rightarrow -\infty}\left(\frac{\frac{9x-x^{2}}{8x-8}}{x}\right)\end{align}\) - step2: Calculate: \(\begin{align}&-\frac{1}{8}\\&-\frac{1}{8}\end{align}\) - step3: Evaluate the limit \(\lim _{x\rightarrow +\infty}\left(r\left(x\right)-ax\right):\) \(\begin{align}&\lim _{x\rightarrow +\infty}\left(\frac{9x-x^{2}}{8x-8}+\frac{1}{8}x\right)\\&\lim _{x\rightarrow -\infty}\left(\frac{9x-x^{2}}{8x-8}+\frac{1}{8}x\right)\end{align}\) - step4: Calculate: \(\begin{align}&1\\&1\end{align}\) - step5: Write the oblique asymptote according to limits value: \(\begin{align}&r\left(x\right)=-\frac{1}{8}x+1\textrm{ }\textrm{is an oblique asymptote}\\&r\left(x\right)=-\frac{1}{8}x+1\textrm{ }\textrm{is an oblique asymptote}\end{align}\) - step6: The function has one oblique asymptote: \(r\left(x\right)=-\frac{1}{8}x+1\) The results for the asymptotes of the function \( r(x) = \frac{9x - x^2}{8x - 8} \) are as follows: 1. **Slant Asymptote**: The slant (or oblique) asymptote is given by the equation: \[ r(x) = -\frac{1}{8}x + 1 \] 2. **Vertical Asymptote**: The vertical asymptote occurs at: \[ x = 1 \] Thus, the slant asymptote is \( -\frac{1}{8}x + 1 \) and the vertical asymptote is \( x = 1 \).