\begin{frame}
  \frametitle{Slant Asymptotes}

  \begin{exampleblock}{}
    Sketch the graph of $f(x) = \frac{x^3}{2x^2+1}$.
    \pause\medskip
    
    \only<-17>{
    The domain is \pause $(-\infty,\infty)$
    \pause\medskip
    
    The \quad $f(0) = \pause 0$ \quad\pause and \quad $f(x) = 0 \;\iff\; \pause x = 0$
    \pause\medskip
    
    Vertical asymptotes: \pause none. \quad \pause Horizontal asymptotes: \pause none
    \pause\medskip
    
    Slant asymptotes: \pause \quad $y = \frac{1}{2}x$ \quad\pause since
    \begin{talign}
      \lim_{x\to \infty} \left(\frac{x^3}{2x^2+1} - \frac{x}{2}\right) 
      &\mpause[1]{= \lim_{x\to \infty} \left(\frac{2x^3 - x(2x^2+1)}{2(2x^2+1)}\right) }\\
      &\mpause[2]{= \lim_{x\to \infty} \left(\frac{-x}{2(2x^2+1)}\right) }
      \mpause[3]{= 0 }
    \end{talign}
    }
    \only<18->{
    \pause[18]\vspace{-2.5ex}
    
    \begin{talign}
      f'(x) = \mpause[1]{ \frac{3x^2(2x^2+1) - x^3(4x)}{(2x^2+1)^2} }
      \mpause[2]{ = \frac{2x^4 + 3x^2}{(2x^2+1)^2} }
      \mpause[3]{ = \frac{x^2(2x^2 + 3)}{(2x^2+1)^2} }
    \end{talign}
    \pause\pause\pause\pause
    Thus $f'(x) > 0$ for all $x\ne 0$. \pause Hence increasing on $(-\infty,\infty)$.
    \pause\medskip
    
    Local minima, maxima: \pause none (since $f'$ does not change sign)
    \pause\medskip

    We have\vspace{-3ex}
    \begin{talign}
      f''(x) = %&= \mpause[1]{ \frac{(8x^3 + 6x)(2x^2+1)^2 - (2x^4 +3x^2)2(2x^2+1)4x}{(2x^2+1)^4} } \\
      \mpause[1]{ -\frac{2x(2x^2 - 3)}{(2x^2+1)^3} }
    \end{talign}\vspace{-2ex}
    \pause\pause
    
    Thus \quad $f''(x) = 0 \;\iff\;$ \quad\pause $x = 0$ \quad or \quad $x = \pm \sqrt{3/2}$
    \begin{tabular}{|c|c|l|}
      \hline
      Interval & $f''(x)$ & \\
      \hline
      \mpause[1]{ $x < -\sqrt{3/2}$ } & \mpause[5]{+} & \mpause{ concave up on $(-\infty,-\sqrt{3/2})$ } \\
      \hline
      \mpause[2]{ $-\sqrt{3/2} < x < 0$ } & \mpause[7]{-} & \mpause{ concave down on $(-\sqrt{3/2},0)$ } \\
      \hline
      \mpause[3]{ $0 < x < \sqrt{3/2}$ } & \mpause[9]{+} & \mpause{ concave up on $(0,\sqrt{3/2})$ } \\
      \hline
      \mpause[4]{ $\sqrt{3/2} < x$ } & \mpause[11]{-} & \mpause{ concave up down $(\sqrt{3/2},\infty)$ } \\
      \hline
    \end{tabular}
    \pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause
    Inflection points: \pause $(-\sqrt{\frac{3}{2}},-\frac{3}{8}\sqrt{\frac{3}{2}})$, $(0,0)$ and $(\sqrt{\frac{3}{2}},\frac{3}{8}\sqrt{\frac{3}{2}})$
    }
    
  \end{exampleblock}
  \vspace{10cm}  
\end{frame}