What is the equation of the line tangent to # f(x)=(-x^2-1)/(x^2+4) # at # x=-3 #?

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Answer 1 · 2018-06-04T22:20:49

The tangent line at that point is #y+10/13=126/169(x+3)#.

We need to find the slope of the tangent line at this point, #x=-3#. To do so, we find #f'(x)# and plug -3 in:

#f'(x)=(-2x(x^2+4)-2x(x^2-1))/(x^2+4)^2=(-2x^3-8x-2x^3+2x)/(x^+4)^2=(-4x^3-6x)/(x^2+4)^2#

#f'(-3)=((-4*-27)-(6*-3))/(9+4)^2=126/169#

We now find the #y#-value at #x=-3#:

#f(-3)=(-9-1)/(9+4)=-10/13#

Using point-slope form, we have the equation of the tangent line as

#y+10/13=126/169(x+3)#