What is the equation of the line tangent to #f(x)= -1/(5x-x^2) # at #x=-3#?

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Answer 1 · 2016-01-13T09:25:18

Tangent Line
#y=11/576 x +19/192#

#f(x) = -1/(5x-x^2#

determine the point of tangency

#f(-3)=-1/(5(-3)-(-3)^2#

#f(-3)=1/24#

Determine the slope #m#

#m=f'(x)=(5-2x)/(5x-x^2)^2#

#m=(5-2(-3))/(5(-3)-(-3)^2)^2#

#m=11/576#

Use point-slope form
#y-y_1=m(x-x_1)#

#y-1/24=11/576(x-(-3))#

#576y-24=11x+33#

#y=11/576 x+57/576#

#y=11/576 x + 19/192#