## Calculus, 10th Edition (Anton)

$$y = - \frac{1}{2}x + \frac{{\ln 2}}{2} - 2$$
\eqalign{ & {\text{Let }}f\left( x \right) = \ln \left| x \right|{\text{ and the point }}{x_0} = - 2 \cr & {\text{then }}f\left( {{x_0}} \right) = f\left( { - 2} \right) = \ln \left| { - 2} \right| = \ln 2 \cr & {\text{we have the point}}\left( { - 2,\ln 2} \right) \cr & {\text{Find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \left( {\ln \left| x \right|} \right)' \cr & f'\left( x \right) = \frac{1}{{\left| x \right|}},{\text{ }}x{\text{ > 0}},\,\,\,\, - \frac{1}{{\left| x \right|}},{\text{ }}x < {\text{0}} \cr & {\text{evaluate }}f'\left( x \right){\text{ at the point }}{x_0}{\text{ to find the slope}} \cr & f'\left( { - 2} \right) = - \frac{1}{{\left| { - 2} \right|}} \cr & f'\left( { - 2} \right) = -\frac{1}{2} \cr & {\text{then }}m = - \frac{1}{2} \cr & \cr & {\text{using the equation of the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right),{\text{ point }}\left( -2, \ln2 \right) \cr & y - \left( { - 2} \right) = - \frac{1}{2}\left( {x - \ln 2} \right) \cr & {\text{Simplify}} \cr & y + 2 = - \frac{1}{2}x + \frac{{\ln 2}}{2} \cr & y = - \frac{1}{2}x + \frac{{\ln 2}}{2} - 2 \cr}