To find the equation of the line tangent to [math]f\left(x\right)=\sqrt{x+3}[/math] at [math]x=13[/math], we need:[br][list][*]The slope of the secant line, [math]m_{tan}[/math], given by [math]f'\left(13\right)[/math]. [/*][*]The ordered pair [math]\left(13,f\left(13\right)\right)[/math].[/*][/list][br][u][b]The Plan[br][/b][/u][list=1][*]To find [math]f'\left(13\right)[/math], we will first find [math]f'\left(x\right)[/math] using the formula, [math]f'\left(x\right)=\lim_{h\rightarrow0}[/math][math]\frac{f\left(x+h\right)-f\left(x\right)}{h}[/math]. Then, we evaluate [math]f'\left(x\right)[/math] at [math]x=13[/math].[/*][*]To find the ordered pair, we plug [math]x=13[/math] into our equation, [math]f\left(x\right)=\sqrt{x+3}[/math].[/*][/list]

[center][/center]From the graph above, we see [math]m_{tan}=\frac{1}{8}[/math]. This is shown algebraically in the [i]Solution[/i] pdf at the end of the page.[br][br]Using point slope form, our equation is given by, [math]\left(y-f\left(a\right)\right)=m_{tan}\left(x-a\right)[/math]. In this case, [math]a=13[/math] and [math]f\left(a\right)=4[/math]. [br][math]y-4=\frac{1}{8}\left(x-13\right)[/math][br][math]y-4=\frac{1}{8}x-\frac{13}{8}[/math][br][math]y-\frac{32}{8}=\frac{1}{10}x-\frac{13}{10}[/math][br][math]y=\frac{1}{8}x+\frac{19}{8}[/math].