\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}