\begin{frame} \frametitle{2nd Midterm Exam - Review} \begin{exampleblock}{} Let $f(x) = x^2$. Find $a > 0$ such that the tangent to the curve at the point $(a,f(a))$ passes through the point $(1,-3)$. \pause\bigskip We have $f'(x) = 2x$. \pause Thus the tangent through $(a,f(a))$ is \begin{align*} y - f(a) = f'(a)\cdot (x - a) \mpause[1]{\quad\implies\quad y - a^2 = 2a(x - a)} \end{align*} \pause We want that the tangent passes through $(1,-3)$\pause, thus \begin{align*} &(-3) - a^2 = 2a(1 - a)\\ &\mpause[1]{\implies\quad (-3) - a^2 = 2a - 2a^2}\\ &\mpause[2]{\implies\quad a^2 - 2a -3 = 0}\\ &\mpause[3]{\implies\quad a = 1 \pm \sqrt{1 + 3}} \end{align*} \pause\pause\pause\pause Thus $a = 3$ \quad\textcolor{gray}{(recall that we were searching for $a > 0$)}\\\pause The tangent is \quad $y - 9 = 6(x - 3)$ \end{exampleblock} \end{frame}