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Can someone help me find the equation of a tangent line of y?

Find the equation of the tangent line to the graph of y=f(x) at x
f(x)=ln(-x) x=-e^3

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Xavier H. | Excellent Math and Science TutorExcellent Math and Science Tutor
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To find the equation of a tangent line, we will need the slope m and a point (xo, yo).  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 (-e3,f(-e3)).
 
Finding the slope: m = f'(-e3)  
 
f(x) = ln(-x)
f(x) = ln(x) + ln(-1)       (property of logs)
 
f'(x) = 1/x
f'(-e3) = -1/e3
 
Finding the point (-e3,f(-e3))
 
f(x) = ln(-x)
f(-e3) = ln(-(e3))
f(-e3) = 3                  (property of logs)
 
Our point is (-e3, 3)
 
Find the equation of a line 
y - yo = m(x - xo)
y = mx - mxo + yo
y = -x/e3 - (-1/e3)*(-e3) + 3
y = -x/e3 - 1 + 3
 
y = -x/e3 - 2      <==== equation of a tangent line