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\begin{frame}
  \frametitle{Exercises}
  
  We know $c = 490$ and
  \begin{talign}
    (1) \quad &h(2) = 2^2a + 2b + 490 = 472\\
    (2) \quad &h(4) = 4^2a + 4b + 490 = 414
  \end{talign}
  \pause\vspace{-1ex}
  
  We simplify
  \begin{talign}
    (1) \quad &4a + 2b + 18 = 0\\
    (2) \quad &16a + 4b + 76 = 0
  \end{talign}
  \pause\vspace{-1ex}
  
  We solve by taking $(2) - 2\cdot (1)$:
  \begin{talign}
    &h(2) = 8a + 40 = 0 \quad\mpause[1]{\implies 8a = -40 \implies a = -5}
  \end{talign}
  \pause\pause\vspace{-1ex}
  
  We get $b$ by plugging $a = -5$ in $(1)$:
  \begin{talign}
    &4\cdot(-5) + 2b +18 = 0 \quad\mpause[1]{\implies 2b = 2 \implies b = 1}
  \end{talign}
  \pause\pause\vspace{-1ex}
  
  Thus \alert{$h(t) = -5t^2 + t + 490$}.
\end{frame}