\begin{frame}
  \frametitle{2nd Midterm Exam - Review}
  
  \begin{exampleblock}{}
    Where does the normal line to $f(x) = x-x^2$ at point $(1,0)$
    intersect the the curve the second time?
    \pause\bigskip
    
    We have:
    \begin{align*}
      f'(x) = 1 - 2x
      \mpause[1]{\quad\implies\quad f'(1) = -1}
    \end{align*}
    \pause\pause
    The normal line at $(1,0)$ has slope $-\frac{1}{-1} = 1$\pause, and thus is
    \begin{align*}
      y - 0 = 1\cdot (x - 1) &&\mpause[1]{y = x-1}
    \end{align*}
    \pause\pause
    We look for the intersection of $f(x)$ and the normal line:
    \begin{align*}
      x-1 = x-x^2 &\mpause[1]{\quad\implies\quad x^2 - 1 = 0} \\
       &\mpause[2]{\quad\implies\quad (x-1)\cdot (x+1) = 0} 
    \end{align*}
    \pause\pause\pause
    Thus the second intersection is at point $(-1,-2)$.
  \end{exampleblock}
\end{frame}