\begin{frame}
  \frametitle{2nd Midterm Exam - Review}

  \begin{exampleblock}{}
    Let $c > \frac{1}{2}$. How many lines through the point $(0,c)$
    are normal lines to $f(x) = x^2$?
    \pause\bigskip
    
    Let $a$ be arbitrary. We construct the normal line at $(a,f(a))$:\vspace{-1.2ex}
    \begin{align*}
      f'(a) = \mpause[1]{2a}
    \end{align*}
    \pause\pause
    As a consequence the normal line at $(a,f(a))$ for \alert{$a\ne 0$} is:\vspace{-1.2ex}
    \begin{align*}
      y - a^2 = \mpause[1]{-\frac{1}{2a}} (x-a)
    \end{align*}
    \pause\pause
    We check for which $a$ the normal goes through $(0,c)$:
    \begin{align*}
      c - a^2 = -\frac{1}{2a} (0-a)
      \mpause[1]{\quad\implies\quad a^2 = c - \frac{1}{2} }
    \end{align*}
    \pause\pause
    Note that $c - \frac{1}{2} > 0$! \pause Thus there are two solutions for $a$.\\\pause
    Note that the normal at $(0,0)$ is vertical and goes through $(0,c)$!\\\pause
    Hence there are three normal lines going through $(0,c)$.    
  \end{exampleblock}
\end{frame}