To find the equation of a tangent line, we will need the slope m and a point (x_{o}, y_{o}). Then, we will use any method to find an equation of a line (i.e. point-slope form, y-intercept form). Here, we will have to find the slope and
then the point (-e^{3},f(-e3)).
Finding the slope: m = f'(-e^{3})
f(x) = ln(-x)
f(x) = ln(x) + ln(-1) (property of logs)
f'(x) = 1/x
f'(-e^{3}) = -1/e^{3}
Finding the point (-e^{3},f(-e^{3}))
f(x) = ln(-x)
f(-e^{3}) = ln(-(e^{3}))
f(-e^{3}) = 3 (property of logs)
Our point is (-e^{3}, 3)
Find the equation of a line
y - y_{o} = m(x - x_{o})
y = mx - mx_{o} + y_{o}
y = -x/e^{3} - (-1/e^{3})*(-e^{3}) + 3
y = -x/e^{3} - 1 + 3
y = -x/e^{3} - 2 <==== equation of a tangent line