What is the equation of the line tangent to # f(x)=x/(x^2-4) # at # x=-1 #?

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Answer 1 · 2015-12-04T03:03:28

#y=-5/9x-2/9#

According to the Quotient Rule:

#f'(x)=((x^2-4)d/dx[x]-xd/dx[x^2-4])/(x^2-4)^2#

#f'(x)=((x^2-4)(1)-x(2x))/(x^2-4)^2#

#f'(x)=(x^2-4-2x^2)/(x^2-4)^2#

#f'(x)=-(x^2+4)/(x^2-4)^2#

To find the tangent line at #x=-1#, determine what point the tangent will intersect:

#f(-1)=(-1)/((-1)^2-4)=1/3#

The tangent line will have a slope of #f'(-1)# at the point #(-1,1/3)#.

#f'(-1)=-((-1)^2+4)/((-1)^2-4)^2=-5/9#

Write the equation in point-slope form: #y-1/3=-5/9(x+1)#

In slope-intercept form: #y=-5/9x-2/9#