#math/calculus If $f$ is differentiable at $x_{0}$, the line tangent to $y=f(x)$ at $(x_{0},f(x_{0}))$ has slope $f'(x_{0})$. By [[0R4 - Lines|point-slope form]], $ y-f(x_{0})=f'(x_{0})(x-x_{0}), $ or equivalently $y=f(x_{0})+f'(x_{0})(x-x_{0})$. ```tikz \begin{document} \begin{tikzpicture}[scale=1.05,line cap=round] \draw[->,black!45] (-0.2,0) -- (5.1,0) node[right] {$x$}; \draw[->,black!45] (0,-0.3) -- (0,2.3) node[above] {$y$}; \draw[very thick,blue,domain=0.2:4.8,samples=100] plot (\x,{0.22*(\x-2.4)*(\x-2.4)+0.7}); \draw[thick,orange!85!black] (0.2,1.545) -- (3.7,-0.303) node[below right] {tangent}; \draw[dashed,black!40] (1.2,0) -- (1.2,1.017); \fill (1.2,1.017) circle (1.8pt) node[above right] {$(x_0,f(x_0))$}; \node[below] at (1.2,0) {$x_0$}; \end{tikzpicture} \end{document} ``` > [!Insight] > The tangent line records the function's value and its instantaneous rate of change at one point. Those two pieces of information produce the best nearby linear model.