To find the equation of the line tangent to [math]f[/math] at the point where [math]x=2[/math], we need the slope of the tangent line, [math]m_{tan}[/math], and one point on the line- in this case, our point will be at [math]\left(2,f\left(2\right)\right)[/math].

To find the point we will be using, we find [math]\left(2,f\left(2\right)\right)[/math]. This is done algebraically in the [i]Solution[/i] pdf at the end of the page. Graphically, we see that the point is [math]\left(2,22\right)[/math].

To find the slope of the tangent line, we will use [math]m_{tan}=\lim_{h\rightarrow0}[/math][math]\frac{f\left(x+h\right)-f\left(x\right)}{h}[/math]. This is done algebraically in the [i]Solution[/i] pdf at the end of the page. We will use a graph to determine the slope of the tangent line:

From the graph above, we can see that as h nears 0, m nears 24. Therefore, [math]m_{tan}=24.[/math]

To create the equation of the line whose slope is 24, and contains the point [math]\left(2,22\right)[/math], we use point-slope form, plugging into: [math]\left(y-f\left(a\right)\right)=m_{tan}\left(x-a\right)[/math].