\begin{frame} \frametitle{The Tangent: Example} \begin{exampleblock}{} Find the equation for the tangent to the curve $x^2$ at point $(1,1)$. \end{exampleblock} \vspace{-.5ex} \begin{itemize} \item<2-> We need to know the slope $m$ of $x^2$ at point $P = (1,1)$. \item<3-> Take point $Q = (x,x^2)$ with $Q \ne P$ to compute the slope. \end{itemize} \medskip \begin{minipage}{.49\textwidth} \scalebox{.7}{ \begin{tikzpicture}[default] \diagram{-1}{5}{-1}{5}{1} \diagramannotate \draw[name path=x2,ultra thick,cgreen] plot[smooth,domain=-1:2.3,samples=20] function{x**2} node[right] {$f(x) =x^2$}; \node[dot] (P) at (1,{1^2}) {}; \node[anchor=north west,at=(P.south)] {$P$}; \onslide<18->{\tangent[ultra thick]{55}{80}{\x^2}{1}} \foreach \s/\x in {4/2,5/2,6/2,7/2,8/2,9/2,10/1.5,11/1.5,12/1.1,13/1.1,14/1.01,15/1.01,16/1.01,17/1.01} { \only<\s>{ \node[dot] (Q) at (\x,{\x^2}) {}; \node[anchor=south east,at=(Q.north)] {$Q$}; \through[red,ultra thick]{2cm}{(1cm+(2-\x)*3cm)}{P}{Q} } } \end{tikzpicture} } \end{minipage} % \pause\pause\pause\pause% \begin{minipage}{.49\textwidth} % Take $Q = (x,x^2)$ with $Q \ne P$. The slope from $P$ to $Q$ is: \begin{talign} m_{PQ} = \frac{Q_y - P_y}{Q_x - P_x} \onslide<6->{= \frac{x^2 - 1}{x - 1}} \end{talign} \pause\pause \begin{tabular}{|l|l|l|l|l|} \hline $x$ & \onslide<8->{$2$} & \onslide<10->{$1.5$} & \onslide<12->{$1.1$} & \onslide<14->{$1.01$} \\ \hline $m_{PQ}$ & \onslide<9->{$3$} & \onslide<11->{$2.5$} & \onslide<13->{$2.1$} & \onslide<15->{$2.01$} \\ \hline \end{tabular}\pause[16]~\ldots \bigskip\pause The closer $Q$ to $P$, the closer $m_{PQ}$ gets to $2$. \pause \structure{Suggests that in $P$ the slope $m = 2$.} \end{minipage} \bigskip\pause Thus the \alert{tangent is $y - 1 = 2(x - 1)$} \pause or \alert{$y = 2x - 1$}. \bigskip \end{frame}